Solution: Benzene is a planar atom with electrons delocalized under or more the ring plane. Henceforth, it is a material wealthy in electrons. As an outcome, electron-inadequate species, i.e.,...
Solution: Acidic person of an animal groups is characterized based on the simplicity with which it can lose its H–iotas. The hybridization condition of carbon in the given compound is: As the s –...
Solution: The expansion of HBr to propene is an illustration of an electrophilic replacement response. Being a corrosive, the hydrogen bromide gives an electrophile, H+. This electrophile assaults...
Solution: Alkanes experience Van-der Waals powers between atoms. The higher the alkane's force, the more prominent is the limit. As the atom stretching expands, the surface region diminishes which...
Solution: Essential carbon particles (1o): Carbon iotas clung to a solitary molecule of carbon are called essential iotas of carbon. The game plan given has comparing five 1o carbon iotas and...
Solution: Benzene changed over to acetophenone
Solution: Benzene changed over to p – nitrotoluene
Solution: Benzene changed over to m-nitrochlorobenzene
Solution: Benzene changed over to p – nitrobromobenzene
Solution: Cyclo-octatetraene isn't planar yet is tub-molded. It is, in this manner, a non-planar framework having 8 n-electrons. In this manner, the atom isn't sweet-smelling as it doesn't contain a...
Solution: Because of the presence of sp3–hybridized carbon, the framework isn't planar. Further, it contains just four n-electrons, subsequently, the framework isn't fragrant on the grounds that it...
Solution: Because of the presence of a sp3-hybridized carbon, the framework isn't planar. It contains six n-electrons yet the framework isn't completely formed since all the six n-electrons don't...
Solution: The fundamental conditions for any fragrant framework are as per the following: (I) Firstly, the compound course of action or construction ought to be planar. (ii) The n-electrons are...
Solution: Benzene is a mixture of the resounding designs and it is displayed as: Every one of the six carbon molecules in benzene are hybridized to sp2.In benzene, every carbon particle's two...
Solution: Hex-2-ene is addressed as displayed beneath: Mathematical isomers of hex-2-ene are as per the following: A cis intensifies dipole second is equivalent to the amount of the C – CH3...
Solution: Burning responses might be characterized as an oxygen or oxygen response of a compound.
Solution: Burning responses might be characterized as an oxygen or oxygen response of a compound.
Solution: Burning responses might be characterized as an oxygen or oxygen response of a compound.
Solution: Burning responses might be characterized as an oxygen or oxygen response of a compound.
Solution: From the given data, the two ozonolysis results of an alkene are pentan-3-one and propanal. Assume the alkene given is A. The converse of the ozonolysis response is the thing that we get,...
Solution: From the data given it gives two moles of an aldehyde of molar mass 44 u when ' A ' goes through ozonolysis. The arrangement of an aldehyde's two moles recommends that the presence of...
Throughout ozonolysis, an ozonide is created as a halfway that has a cyclic construction; it goes through cleavage to give the eventual outcomes. Ethanal and pentan-3-one are acquired from the...
Solution: Ozonolysis of 1 – Phenylbut-1-ene is displayed as The resulting IUPAC names of the items are : Item (I) benzaldehyde, and Item (II) propanal
Solution: Ozonolysis of 2-Ethylbut-1-ene is displayed as: The resulting IUPAC names of the items are: Item (I) pentan-3-one, and Item (II) methanal
Solution: Ozonolysis of 3, 4-Dimethylhept-3-ene is displayed as: The item names resulting to IUPAC are: Item (I) butan-2-one, and Item (II) Pentan-2-one
Solution: Ozonolysis of Pent-2-ene is displayed as: The accompanying Product IUPAC names are: Item (I) ethanal, and Item (II) propanal.
Solution: The resulting primary isomers are likely for C5H8 with one triple bond: The IUPAC name of Compound (I) is Pent-1-yne, Compound (II) is Pent-2-yne, and Compound (III) is...
Solution: 4-Ethyldeca-1, 5, 8-triene is the necessary IUPAC name
Solution: 5-(2-Methylpropyl)- decane is the necessary IUPAC name
Solution: 2-Methyl phenol is the necessary IUPAC name
Solution: 4-Phenyl but-1-ene is the necessary IUPAC name
Solution: 1, 3-Butadiene or Buta-1,3-diene is the necessary IUPAC name
Solution: Pen-1-ene-3-yne is the necessary IUPAC name
![\[\mathbf{C}{{\mathbf{H}}_{\mathbf{3}}}~\mathbf{CH}\text{ }=\text{ }\mathbf{C}\text{ }{{(\mathbf{C}{{\mathbf{H}}_{\mathbf{3}}})}_{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c63b9b35d0c7e5500238909182cb534b_l3.png "Rendered by QuickLaTeX.com")
Solution: 2-Methylbut-2-ene is the necessary IUPAC name
Solution: The methane chlorination measure works through a free extreme chain instrument. Stage 1: Initiation: The outcome starts with the hemolytic cleavage of Cl – Cl bond as: Stage 2:...
and 
![\[BF _{3}+ NH _{3} \rightarrow F _{3} B \cdot NH _{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f87a2a8a2c94b67d7949b58594c840a9_l3.png "Rendered by QuickLaTeX.com")
Answer: F3B.NH3 is a reaction product formed by the reaction of BF3 + NH3. As a result of changing the hybridization of the B-atom to sp3, the product NH3 is produced. The hybridization of N-atoms,...
![\[AlCl _{3}+C l^{-} \rightarrow A l C l_{4}^{-}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6a3539afdf8fa241be293a500542ae7d_l3.png "Rendered by QuickLaTeX.com")
Answer: The hybridization of Al atom alters from $sp ^{2}$ in $AlCl _{3}$ to $sp ^{3}$ in $AlCl _{4}^{-}$. In $AlCl _{3}$, Al contains 3 bond pairs of electrons and zero lone pair of electrons....
hybrid orbitals.Answer: Intermixing orbitals of various energies to redistribute their energy and create new orbitals with equal energies and forms. The new orbitals are hybridized. In this way, two s orbitals...
and 
has higher dipole moment and why?Answer: When comparing dipole moments, ammonia (1.47D) has a higher dipole moment than $NF_3$. (0.24D). Both molecules have a pyramidal molecular geometry, which is characteristic of this type of...
i) velocity of the oscillator is zero ii) speed of the oscillator is maximum Answer: (i) The oscillator's velocity is zero when the points A, C, E, and G are at their extreme positions. (ii) The...
Answer: The dipole moment of a molecule with symmetrical and linear geometries is zero because they are vector in nature and the dipole of different bonds cancel with one another, whereas the dipole...
Answer: The square planar geometry has a bond angle of 90 degrees with the equilateral triangle geometry. In comparison to the tetrahedral bond angle, which is 109.5 0, this is less favourable. This...
, and 
.Answer: The differential in electronegativity between the atoms that make up a molecule determines the ionic character of that molecule. As a result, the greater the difference between two...
Answer: When two separate atoms join to form a covalent connection, their electrons are not shared equally. A more electronegative atom's nucleus attracts the bond pair. In this case, an...
Answer: Sr. No Electronegativity Electron affinity 1 The electronegativity of an atom in a chemical compound refers to its proclivity to attract the shared pairs of electrons that are present in the...
Answer: The difference in electro-negativities of atom components causes polarisation in heteronuclear molecules. So one end gets a positive charge and the other gets a negative charge. Molecules...
can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing ? If not, give reasons for the same.Answer: The positions of the atoms remain constant in the canonical forms, while the positions of the electrons change. The positions of atoms shift in the specified canonical forms. As a result,...
ion.Answer: Although the properties of the carbonate ion cannot be described by a single structure, they can be described by two or more resonance structures. The carbonate ion's true structure is a...
Answer: “The equilibrium distance between the nuclei of two bound atoms in a molecule is defined as bond length.”
Answer: During the formation of a molecule, the extent of bonding that happens between two atoms is represented by the bond strength of the molecule. As the bond strength develops, the bond becomes...
Answer: Ammonia's central atom (N) has 1 lone pair and 3 bond pairs. Water's central atom (O) has two lone pairs and two bond pairs. The two lone pairs on the O- atom of water reject the two bond...
Answer: An ionic link is formed by moving electrons from one atom to another. This means that ionic bond formation is dependent on neutral atom flexibility. The lattice energy of the molecule...
Answer: "Atoms can combine either by transferring or sharing valence electrons to achieve the closest inert gas configuration," says the Octet Rule. Octet rule describes how chemical bonds occur...
Solution: Ne displays neither negative nor positive oxidation no. That is 0.
Solution: I displays both negative and positive oxidation no. That is - 1, +1, +3, +5 and +7.
Solution: Cs displays just certain oxidation no. That is +1.
Solution: F exhibits only negative oxidation number. That is -1.
Solution: $\mathrm{KAl}\left(\underline{\mathrm{S}} \mathrm{O}_{4}\right)_{2} .12 \mathrm{H}_{2} \mathrm{O}$ Let expect oxidation number of S is x. \[\begin{array}{*{35}{l}}...
Solution: $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$ Let accept oxidation number of S is x. \[\begin{array}{*{35}{l}} Oxidation\text{ }number\text{ }of\text{ }O\text{ }=\text{...
Solution: NaBH4 Let accept oxidation number of B is x. \[\begin{array}{*{35}{l}} Oxidation\text{ }number\text{ }of\text{ }Na\text{ }=\text{ }+1 \\ ~ \\ Oxidation\text{ }number\text{...
Solution: CaO2 Let expect oxidation number of O is x. \[Oxidation\text{ }number\text{ }of\text{ }Ca\text{ }=\text{ }+2\] Then, at that point, we have: \[\begin{array}{*{35}{l}}...
Solution: K2MnO4 Let expect oxidation number of Mn is x. \[\begin{array}{*{35}{l}} Oxidation\text{ }number\text{ }of\text{ }K\text{ }=\text{ }+1 \\ ~ \\ Oxidation\text{ }number\text{...
Solution: H4P2O7 Let expect oxidation number of P is x. \[\begin{array}{*{35}{l}} Oxidation\text{ }number\text{ }of\text{ }H\text{ }=\text{ }+1 \\ ~ \\ Oxidation\text{ }number\text{...
Solution: NaHSO4 Let expect oxidation number of S is x. \[\begin{array}{*{35}{l}} Oxidation\text{ }number\text{ }of\text{ }Na\text{ }=\text{ }+1 \\ ~ \\ Oxidation\text{ }number\text{...
Solution: NaH2PO4 Let expect oxidation number of P is x. We realize that, \[\begin{array}{*{35}{l}} Oxidation\text{ }number\text{ }of\text{ }Na\text{ }=\text{ }+1 \\ ~ \\...
Solution: (b) Graphite is thermodynamically the most steady type of carbon.
Solution: (c) Boron in diborane is sp3 hybridized.
Solution: Catenation A few components or particles (like carbon) may interface with one another through solid covalent bonds to shape long chains or branches. This trademark is known as catenation....
Solution: Allotropy Allotropy is the presence of a component in more than one structure, having distinctive actual properties yet similar synthetic properties. The various types of a component are...
Solution: Inert pair impact As one actions down the gathering, s-block electrons decline their propensity to take part in substance holding. This impact is known as the dormant pair impact. The...
Solution: Thallium is a piece of the intermittent table gathering 13. For this gathering, the most widely recognized oxidation state is + 3. Heavier individuals from this gathering, nonetheless,...
Solution: Carbon ionizing enthalpy (the principal component in bunch 14) is exceptionally high (1086 kJ/mol). That is normal due to its little size. Nonetheless, there is a sharp decline in enthalpy...
Solution: Aluminum wire is utilized to make transmission links. Aluminium wires are good conductors of electricity and also it is cheap metal easily available, light in weight and also very ductile....
Solution: Aluminum utensils ought not be kept in water for the time being. The oxygen in water responds to make a dainty layer of aluminum oxide with aluminum. This layer keeps further response from...
Solution: Aluminum compounds are utilized to make airplane body. Aluminum has a high protection from elastic and is light. It might likewise be alloyed to various metals like Si, Mg, Cu, Mn and Zn....
Solution: Diamond is utilized as a rough Carbon is sp3 hybridized in Diamond. With the assistance of solid covalent bonds, every carbon molecule is bound to four other carbon iotas. These covalent...
Solution: Graphite is utilized as the grease. Graphite has a layered construction, and the powers of frail van der Waals tie various layers of graphite together. These layers might slide one over...
Solution: A combination of weaken NaOH and aluminum pieces is utilized to open the channel. Sodium hydroxide and aluminum respond to shape aluminate (III) sodium tetra hydroxy and hydrogen gas....
Solution: Conc.HNO3 can be shipped in an aluminum holder. As it responds with aluminum to frame a slight defensive oxide layer on the aluminum surface, concentrated HNO3 can be put away and shipped...
Solution: Carbon dioxide is a gas that is vital for our endurance. The expanded CO2 content in the climate, nonetheless, represents a genuine danger. An increment in petroleum product ignition,...
Solution: Given its capacity to frame a complex with hemoglobin, carbon monoxide is profoundly toxic. The previous hinders restricting with oxygen by Hb. Subsequently, an individual passes on...
Solution: Chromatography It is generally utilized for the partition and cleansing of natural mixtures. Rule: The guideline on which it works is that singular parts of a blend move at various speeds...
Solution: Distillation This strategy is utilized to isolate non-unpredictable fluids from unstable pollutants. It is likewise utilized when the parts have an extensive distinction in their edges of...
Solution: Crystallization Crystallization is utilized to decontaminate strong natural mixtures. Standard: The guideline on which it works is the distinction in the dissolvability of the compound and...
Solution: Substitution (nucleophilic) response since modification of particles happens.
Solution: Elimination (bimolecular) response since response hydrogen and bromine are taken out to shape ethene.
Solution: Addition (electrophilic) response since two reactant particles joins to shape a solitary item.
Solution: Substitution (nucleophilic) response since bromine bunch gets subbed by – SH bunch.
Solution: A nucleophile is a reagent that has an electron pair and will give it away. It is otherwise called a core adoring reagent. An electrophile is a reagent which needs an electron pair and is...
Solution: A nucleophile is a reagent that has an electron pair and will give it away. It is otherwise called a core adoring reagent. An electrophile is a reagent which needs an electron pair and is...
Solution: A nucleophile is a reagent that has an electron pair and will give it away. It is otherwise called a core adoring reagent. An electrophile is a reagent which needs an electron pair and is...
Solution: A nucleophile is a reagent that has an electron pair and will give it away. It is otherwise called a core cherishing reagent. Ex: NC–, OH–, R3C–(carbanions), and so on An electrophile is a...
Solution: Since NO2 has a place with the electron-pulling out bunch, it shows – I impact. NO2 attempts to diminish the negative charge on the compound by pulling out the electrons toward it. This...
Solution: The initial five individuals from each homologous series starting with the given mixtures are H–CH=CH2: Ethene CH3–CH=CH2: Propene CH3–CH2–CH=CH2: 1-Butene CH3–CH2–CH2–CH=CH2: 1-Pentene...
Solution: CH3COCH3: Propanone CH3COCH2CH3: Butanone CH3COCH2CH2CH3 : Pentan-2-one CH3COCH2CH2CH2CH3: Hexan-2-one CH3COCH2CH2CH2CH2CH3 : Heptan-2-one
Solution: The initial five individuals from each homologous series starting with the given mixtures are H–COOH: Methanoic corrosive CH3–COOH: Ethanoic corrosive CH3–CH2–COOH: Propanoic corrosive...
Solution: Out of the two utilitarian gatherings present in the given compound, the alcoholic gathering is the central practical gathering. Consequently, the parent chain will have an – ol postfix....
Solution: If the substituents in the chain are in identical positions, then, at that point, the lower number is given to the substituent bunch in sequential request. Along these lines, the right...
Solution: The locant number should begin from the base. Here, 2,4,7 is lower than 2,5,7. Subsequently, the right IUPAC name would be 2,4,7-Trimethyloctane.
solution: The prefix di shows that there are two methyl bunches in the chain. Along these lines, the right IUPAC name would be 2,2-Dimethylpentane.
![\[C6H6\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-250e7c087123a8d0267c095925b18800_l3.png "Rendered by QuickLaTeX.com")
Solution: C6H6 All the 6 carbon particles in benzene are sp2 hybridized
![\[CH2=CHCN\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-52ef84ac98a604743dfb23732dfbb165_l3.png "Rendered by QuickLaTeX.com")
Solution: C-1 is sp2 hybridized. C-2 is sp2 hybridized. C-3 is sp hybridized.
![\[\left( CH3 \right)2CO\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a2bea77220b7e99546204fa04a862cfa_l3.png "Rendered by QuickLaTeX.com")
Solution: C - 1 and C-3 are sp3 hybridized. C-2 is sp2 hybridized.
![\[CH3CH=CH2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6a2ef16862caa7e37fd3861c7df53de4_l3.png "Rendered by QuickLaTeX.com")
Solution: C-1 is sp3 hybridized. C-2 is sp2 hybridized. C-3 is sp2 hybridized.
![\[CH2=C=O\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3ffdae7fcb2301f4db1b66d768c446ab_l3.png "Rendered by QuickLaTeX.com")
Solution: C-1 is sp2 hybridized. C-2 is sp hybridized.
, calculate the energy of the source.Frequency of radiation $\nu$ $\nu =\frac{1}{2.0\times 10^{-9}s}$ $\nu =5.0\times 10^{8}s^{-1}$ Energy (E) of source = Nhν Where, N is the no. photons emitted h is Planck’s constant ν denotes the...
The energy associated with an incident photon (E) must equal the sum of its kinetic energy and the work function (W0) of the radiation, according to the rule of conservation of energy. E = W0 + K.E...
, calculate the energy with which it is bound to the nucleus.Energy of incident photon (E) is given by, $E=\frac{hc}{\lambda }$ $E=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(150\times 10^{-12})}=1.3252\times 10^{-15}\, J$ $\simeq 13.252\times 10^{-16}J$...
(Hz) [1/3^2 – 1/n^ 2 ] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.Wavelength of the transition = 1285 nm =$1285 \times 10^{-9} m$(Given) $\nu =3.29\times 10^{15}(\frac{1}{3^{2}}-\frac{1}{n^{2}})$ Since$ \nu =\frac{c}{\lambda }$ =$\frac{3\times...
The radius of the n th orbit of hydrogen-like particles is given by, $r=\frac{0.529n^{2}}{Z}$ $r=\frac{5.29n^{2}}{Z}pm$ For radius (r1) = 1.3225 nm $=1.32225\times 10^{-9}m$ $=1322.25\times...
, calculate de Broglie wavelength associated with this electron.As per de Broglie’s equation, $\lambda =\frac{h}{m\nu }$ =$\frac{(6.626\times 10^{-34})}{9.103939\times 10^{-31}kg(1.6\times 10^{6}ms^{-1})}$ =$4.55\times 10^{-10}m\lambda =455pm$ Therefore, de...
From de Broglie’s equation, $\lambda =\frac{h}{m\nu }$ $\nu=\frac{h}{m\lambda}$ Where, v denotes the velocity of the neutron h is Planck’s constant m is the mass of the neutron λ is the wavelength...
, calculate the de Broglie wavelength associated with it.As per de Broglie’s equation, $\lambda =\frac{h}{m\nu}$ Where, λ is the wavelength of the electron h is Planck’s constant m is the mass of the electron v denotes the velocity of electron...
. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.As per de Broglie’s expression, $\lambda =\frac{h}{m\nu }$ $ =\frac{(6.626\times 10^{-34})}{0.1kg(4.37\times 10^{5}ms^{-1})}$=$1.516\times 10^{-38}m$
As per Heisenberg’s uncertainty principle, ∆x.∆p >= h/4π Where, ∆x = uncertainty in the position of the electron ∆p = uncertainty in the momentum of the electron Substituting the given values in...
The nuclear charge that electrons (which are present in atoms containing multiple electrons) feel is determined by the distance between their orbital and the atom's nucleus. The smaller the...
The net positive charge acting on an electron in an atom's orbital with more than one electron is known as the nuclear charge. The distance between the orbital and the nucleus is inversely...
(a)Phosphorus (P): The atomic number of phosphorus is 15 Electronic configuration of Phosphorus: 1s 2 2s 2 2p 6 3s 2 3p 3 This can be represented as follows: From the diagram, it can be observed...
(a)n = 4 (Given) For some value of ‘n’, the values of ‘l’ range from 0 to (n – 1). Here, the possible values of l are 0, 1, 2, and 3 Therefore, a total of 4 subshells are possible when n=4: the s,...
To avoid odours and insects, the compost pit should be covered. To avoid interfering with the breakdown of the wastes, non-biodegradable wastes should not be thrown into the compost pit. They should...
To begin, trash must be separated into biodegradable and non-biodegradable categories. Biodegradable materials, such as leaves and food wastes, are deposited in landfills alongside microorganisms...
Bacteria eat phytoplankton, and this process necessitates the presence of dissolved oxygen. As a result, the greater the number of Phytoplankton present, the greater the consumption of dissolved...
The greenhouse gases in our atmosphere capture the sun's UV rays and heat up the earth. Without greenhouse gases, the earth will be unable to retain any heat, which is necessary for the existence of...
The manufacturing process employs our current understanding of chemistry principles to produce, develop, and deploy chemical compounds and products that reduce the amount of harmful chemicals in the...
A pesticide is a combination of two or more chemicals that is used to kill bugs. Plant diseases, weeds, insects, mollusks, and other pests that damage plants and cause them to die must all be...
Pesticides and fertilisers are the most common contaminants that pollute soil. Because pesticides like DDT are not soluble in water, they remain in the soil for prolonged periods of time, polluting...
The quantity of oxygen required by bacteria to breakdown organic materials in a given volume of water is known as biochemical oxygen demand. A BOD of less than 5 ppm indicates clean water, but a BOD...
Water pollution is caused by human activities such as storm-water drainage, run-off from agricultural areas, wastewater treatment plant emissions, and so on. Industries and factories emit toxic...
Water pollution happens when undesired and unwanted chemicals are introduced into water bodies, suffocating the aquatic life that lives there. The following are the primary sources of water...
Polar stratospheric clouds offer a surface for the reaction of hypochlorous acid and chlorine nitrate, which produces molecular chlorine after further reaction. Photolysis of HOCl and Molecular...
In the stratosphere, ozone is actually formed by the action of UV rays on Dioxygen molecules(O2). (i) O2(g) →UV O(g) +O(g) (ii) O2(g) + O(g) ↔UV O3(g) The second reaction demonstrates that the...
Photochemical smog is oxidising in nature because it contains NO2 and O3, which cause rubber, stones, metals, and painted surfaces to corrode. Formaldehyde, PAN, and acrolein are also found in...
The interaction of sunlight with nitrogen oxides and hydrocarbons produces photochemical smog. The major components of photochemical smog include formaldehyde, nitric oxide, ozone, PAN, and...
Smog is a combination of smoke and fog which causes air pollution. There are two types of smog: a) Photochemical smog b) Classical smog They can be differentiated as follows: Photochemical smog ...
By interacting with water in the presence of ambient oxygen, nitrogen oxides (NO2, NO) and sulphur oxides (SO2 and SO3) produced by the combustion of coal, car gasoline, and other fossil fuels...
Tropospheric pollution is caused by the presence of undesirable chemicals in the troposphere's lowest layer. Nitrogen oxides, sulphur oxides, carbon, and hydrocarbons are the most common...
The study of biological and chemical processes that occur in nature is referred to as environmental chemistry. It also investigates the chemical species' reactions, origins, impacts, and...
A total number of 15 lines (5 + 4 + 3 + 2 + 1) will be obtained in this hydrogen emission spectrum. The total number of spectral lines emitted when an electron drops to the ground state from the...
The expression for the ionization energy is given by, $E_{n} =\frac{-(2.18\times 10^{-18})Z^{2}}{n^{2}}$ Where Z denotes the atomic number and n is the principal quantum number For the ionization...
The $n_{i} = 4$ to$ n_{f}$= 2 transition results in a spectral line of the Balmer series. The energy involved in this transition can be calculated using the following expression: $E=2.18\times...
Threshold wavelength of the radiation $(\lambda _{0})= 6800 Å=6800\times 10^{-10}\, m$ Threshold frequency of the metal $(\nu _{0}) =\frac{c}{\lambda _{0}}=$ $\frac{3\times 10^{8}ms^{-1}}{6.8\times...
Power of the bulb,$ P = 25 Watt = 25\, Js^{-1}$ Energy (E) of one photon= $h\nu =\frac{hc}{\nu} $ Substituting these values in the expression for E: $E=\frac{(6.626\times 10^{-34})(3\times...
Ionization energy (E) of sodium =$ \frac{N_{A}hc}{\lambda }$ =$\frac{(6.023\times 10^{23}\, mol^{-1})(6.626\times 10^{-34})Js(3\times 10^{8})ms^{-1}}{242\times 10^{-9}m}$ =$4.947\times 10^{5}\, J\,...
m strikes on metal surface, the work function of the metal is 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron 
.(i) Energy of the photon $(E)= h\nu =\frac{hc}{\lambda }$ Where, h denotes Planck’s constant, whose value is $6.626\times 10^{-34}\,Js$ c denotes the speed of light =$ 3\times 10^{8}\,m/s$...
Energy of one photon (E) =$ h\nu$ Energy of ‘n’ photons $E_{n} = nh\nu \Rightarrow n=\frac{E_{n}\lambda }$ Where, \lambdaλ is the wavelength of the photons = 4000 pm = $4000\times 10^{-12}\, m$ c...
Hz. (ii) have a wavelength of 0.50 Å.(i) The energy of a photon (E) can be calculated by using the following expression: $E= h\nu$ Where, ‘h’ denotes Planck’s constant, which is equal to $6.626\times 10^{-34}\, Js\nuν$ (frequency of...
Rearranging the expression, $\lambda =\frac{c}{\nu }$ the following expression can be obtained, $\nu =\frac{c}{ \lambda }$ ……….(1) Here, $\nu$ denotes the frequency of the yellow light c denotes...
(i)\({}_{17}^{35}C\) (ii)\({}_{92}^{233}C\) (iii)\({}_{4}^{9}C\)
(i) 1 molecule of methane contains 10 electrons (6 from carbon, 4 from hydrogen) Therefore, 1 mole of methane contains 10*NA = 6.022*1024 electrons. (ii) Number of neutrons in 14g (1 mol) of 14C =...
produced if 
g 
reacts with 
g of 
? (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass.(i) Balance the given equation: $N_{ 2 }\;(g) \; + \; 3H_{ 2 } \;(g) \; \rightarrow \; 2NH_{ 3 }\;(g) $ Thus, 1 mole (28 g) of N2 reacts with 3 mole (6 g) of H2 to give 2 mole (34 g)...
and 0.50 M different?Molar mass of $Na_{ 2 }CO_{ 3 }$ = (2 × 23) + 12 + (3 × 16) = $106 g mol^{ -1 }$ 1 mole of $Na_{ 2 }CO_{ 3 }$ means 106 g of $Na_{ 2 }CO_{ 3 }$ Therefore, 0.5 mol of $Na_{ 2 }CO_{ 3 }$...
2H2(g) +O2(g) →2H2O(g) 2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of vapour. Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to...
(g)(i) 1 g Au (s) = $\frac{ 1 }{ 197 }$ mol of Au (s) = $\frac{ 6.022 \; \times \; 10^{ 23 } }{ 197 }$ atoms of Au (s) = $3.06 \times \; 10^{ 21 }$ atoms of Au (s) (ii) 1 g Na (s) = $\frac{ 1...
Mole fraction of $C_{ 2 }H_{ 5 }OH$ = $\frac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}$ $0.040 = \frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; +...
atom in g?1 mole of carbon atoms =$ 6.023 \; \times \; 10^{ 23 }$atoms of carbon = 12 g of carbon Therefore, mass of $1 _{}^{ 12 }\textrm{ C }$ atom = $\frac{ 12 \; g }{ 6.022 \; \times \; 10^{ 23 }}$...
(ii) 5 × 5.365 (iii) 0.012 + 0.7864 + 0.0215(i) $\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }$ Least precise no. of calculation = 0.112 Therefore, no. of significant numbers in the answer = No. of significant numbers in...
Molar mass of Argon: =$ [( 35.96755 \; \times \; \frac{ 0.337 }{ 100 }) + ( 37.96272 \; \times \; \frac{ 0.063 }{ 100 })+ ( 39.9624 \; \times \; \frac{ 99.600 }{ 100 })]$ =$ [0.121 + 0.024 + 39.802]...
(i) 52 moles of Ar 1 mole of Ar = $6.023 \; \times \; 10^{ 23 }$atoms of Ar Therefore, 52 mol of Ar = 52 × $6.023 \; \times \; 10^{ 23 }$atoms of Ar = $3.131 \; \times \; 10^{ 25 }$ atoms of Ar...
0.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L of water ≡ $[(0.75 mol)\times(36.5 g mol–1 )]$ HCl is present in 1 L of water ≡ 27.375 g of HCl is present in 1 L of water Thus, 1000 mL of...
1 mol of $MnO_{2}$ = 55 + 2 × 16 = 87 g 4 mol of HCl = 4 × 36.5 = 146 g 1 mol of $MnO_{2}$ reacts with 4 mol of HCl 5 g of $MnO_{ 2 }$will react with: =$ \frac{146 \; g}{87 \; g} \; \times \; 5...
Reagent limitation: It establishes the magnitude of a reaction. It is the first to be consumed in a reaction, causing the process to come to a halt and limiting the number of products produced. (i)...
Time taken = 2 ns = $2 \times10^{ -9 }$ s Now, Speed of light =$3 \times10^{ 8 } ms^{ -1 }$ So, Distance travelled in 2 ns = speed of light * time taken =$(3 \times10^{ 8 })(2 \times10^{ -9 })$ = $6...
(a) If the mass of N2 is set at 28 g, the mass of O2 that will combine with it is 32 grammes, 64 grammes, 32 grammes, and 80 grammes. O2 has a mass-to-number ratio of 1: 2: 1: 5. As a result, the...
(a) The number after round up is: 34.2 (b) The number after round up is: 10.4 (c)The number after round up is: 0.0460 (d)The number after round up is: 2808
(i) 0.0027: 2 significant numbers. (ii) 209: 3 significant numbers. (iii) 6005: 4 significant numbers. (iv) 136,000:3 significant numbers. (v) 900.0: 4 significant numbers. (vi) 2.0035: 5...
(a$) 0.0048= 4.8 \times10^{-3}$ (b) $234,000 = 2.34 \times10^{5}$ (c) $8008= 8.008 \times10^{3}$ (d) $500.0 = 5.000 \times10^{2}$ (e) $6.0012 = 6.0012 \times10^{0}$
(a) $1 ppm = 1 part out of 1 million parts.$ Mass percent of 15 ppm chloroform in H2O= $\frac{15}{{{10}^{6}}}\times 100$ $\approx 1.5 \times{10}^{-3}%$ $ (b)$100 grams of the sample is having 1.5 ×...
The meaningful numbers that are known with certainty are known as significant figures. Significant numbers imply that the tested value is unknown. For example, if the experiment yielded 15.6 mL, 15...
Kilogram is the SI unit of mass (kg) Mass: “Mass is defined as the mass equivalent to the mass of the international kilogramme prototype.”
As per definition, pressure is force per unit area of the surface. $P = \frac{F}{A}$ =$ \frac{1034\;g\;\times \;9.8\;ms^{-2}}{cm^{2}}\times \frac{1\;kg}{1000\;g}\times \frac{(100)^{2}\;cm^{2}}{1...
, what is its volume needed for making 2.5 L of its 0.25 M solution?$Molar mass of CH_{3}OH$ = (1 * 12) + (4 * 1) + (1 * 16) =$32 g mol^{-1}$ = $0.032 kg mol^{-1}$ Molarity of the solution = $\frac{0.793\;kg\;L^{-1}}{0.032\;kg\;mol^{-1} }$ = 24.78 mol{L}^{-1}...
in mol 
if its 20 g are dissolved in enough water to make a final volume up to 2L?Molarity (M) is as given by, = $\frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}$ =$\frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}$ =...
Answer: (b) F > O > Cl > N During a period, as we move from left to right, the non-metallic characteristics of the elements become more prominent. As a result, F > O > N.
Answer: c) F > N > C > B > Si While moving from left to right in a period, the nonmetallic property of the elements reduces as we move from left to right in the period. As a result, F...
Answer: d) K > Mg > Al > B During a period, the metallic nature of the components diminishes as we move from left to right in the period. As a result, Mg > Al. With each group we...
Answer: (d) is a false statement When comparing orbitals with a lower value of 'n' to orbitals with a higher value of 'n,' it is easier to remove an electron from the lower value of 'n' orbital....
Answer: (a) Nuclear charge (Z) Because, in the case of isoelectronic species, the atomic size decreases as the number of nuclear charge increases (Z). e.g. the arrangement according to increasing...
Answer: Option c) Because the nucleus is made up of protons and neutrons, the mass of the nucleus has no effect on the valence shell. Protons, or nuclear charges, have an effect on the valence...
Answer: Correcting the incorrect statement is represented by option (b). It explains that "the d-block has 8 columns because a maximum of 8 electrons can occupy all of the orbitals in a d-subshell,"...
Answer: (c) The period in the Modern periodic table indicates the value of ‘n’ i.e. a principal quantum number. A periodic table's period reflects the value of a particular element's major quantum...
Answer: i) $Li_2O$ is formed when the alkali metal lithium (with one valence electron) and group 16 element oxygen (with two valence electrons) mix. (ii) The alkaline earth metal magnesium (which...
and the second 
ionization enthalpies (in 
) and the 
electron gain enthalpy (in ) of a few elements are given below: Which of the above elements is likely to be : (a) the least reactive element. (b) the most reactive metal. (c) the most reactive non-metal. (d) the least reactive non-metal. (e) the metal which can form a stable binary halide of the formula MX2 (X=halogen). (f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?Elements (\Delta _{i}H_{1})(ΔiH1) (\Delta _{i}H_{2})(ΔiH2) (\Delta _{eg}H)(ΔegH) 1 520 7300 -60 2 419 3051 -48 3 1681 3374 -328 4 1008 1846 -295 5 2372 5251 +48 6 738 1451 -40 Answer:...
for 
(ii) (n1) 
for 
, and (iii) 
for 
, in the periodic table.Answer: i) Since n = 6, the element is in period 6. f-block element because the last electron enters f-orbital. They are in the third group. So, 54 + 7 + 2 + 1 Equals 64. So Gadolinium is required....
Answer: The general outer electronic configuration of s block elements is $ns ^{(1-2)}$. The general outer electronic configuration of p block elements is $ns ^{2} np ^{(1-6)}$. The general outer...
Answer: Group 1 elements have only one valence electron. So they tend to expel this electron to form a stable inert gas. The ionization enthalpies of the elements drop in group 1. Expulsion of the...
Answer: (a) The electrical configuration of an element with 5 electrons in the outer subshell is ns2np5. The electrical configuration of the Halogen group is the same. As a result, the elements...
Answer: Metals lose electrons to produce cations. Metals have low ionization, negative electron, and electronegativities enthalpies. ionic compounds and basic oxides Non-metals readily take...
Answer: The ionization enthalpy of any atom is determined by the number of protons and electrons in the atom. However, the isotopes of any element have the same number of electrons and protons as...
Answer: a) An atom loses one electron when it expels one, but the nuclear charge remains the same. In an atom, electron-electron repulsion reduces. So the effective nuclear charge increases. As a...
Answer: Electronegativity is a property of any element that is subject to change. Electronegativity varies depending on the substance being studied. The following statement is erroneous: "The...
Answer: Electron gain enthalpy Electronegativity It is the common tendency of an atom to attract outside electrons It is the general tendency of an atom to attract shared pair of electrons It is the...
Answer: Oxygen is in p block element group 16. It is the first of 16 members. It is a non-metal found in the earth's crust. When an electron is added to an O atom, it releases energy. O has a...
Answer: i) O and F are from the same period. As the electron is inserted in the same shell, the atomic size of O-atom is larger than F-atom. It has one less proton than F-atom. So an O-atom nucleus...
B Al Ga In Tl 801 577 579 558 589 Answer The ionization enthalpy decreases with group size. B and Al agree. Ga has a higher ionization enthalpy than Al. Tend electrons in the inner shell of...
Answer: The factors that cause the ionization enthalpy to drop as we travel down the main group are as follows: 1. “Inner shells increase in number as we proceed down the group.” The nucleus' inner...
Answer: Because magnesium has a larger 1st ionization enthalpy than sodium, Magnesium has bigger atoms than sodium. A higher effective nuclear charge than sodium, Magnesium So sodium requires less...
Answer: i) The electron that can be evacuated from Be(beryllium) atom is 2s, but the electron that can be expelled from boron is 2p. 2s – electron, and nucleus attract more strongly than 2s –...
. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1.Answer: Given that, the electron of hydrogen is having $-2.18 * 10^{-18} J$ in the ground state. Thus, To eject an electron from the ground state in a $H ($ hydrogen $)-$ atom, a certain amount of...
Answer: In the ground state of an isolated gaseous atom, ionisation enthalpy is required to eject an electron. Even though the atoms are far separated in the gaseous state, there are some attraction...