Solution: The difference in electronegativity between constituent atoms determines the ionic character of a molecule. As a result, the greater the difference, the greater the ionic character of a...
Solution: The bond pair of electrons are not shared equally when two unique atoms with different electronegativities join to form a covalent bond. The bond pair is attracted to the nucleus of an...
Solution: "Electronegativity refers to an atom's ability to attract a bond pair of electrons towards itself in a chemical compound." Sr. No Electronegativity Electron affinity 1 A tendency to...
Solution: There is a difference in electro-negativities of constituents of the atom in a heteronuclear molecule, which causes polarisation. As a result, one end gains a positive charge, while the...
Solution: $CO_2$ has a dipole moment of 0 according to experimental results. And it's only possible if the molecule's shape is linear, because the dipole moments of the C-O bond are equal and...
Solution: Below is a list of Lewis symbols. To form a cation, a metal atom loses one or more electrons, while a nonmetal atom gains one or more electrons. Ionic bonds are formed between cations and...
Solution: Below is a list of Lewis symbols. To form a cation, a metal atom loses one or more electrons, while a nonmetal atom gains one or more electrons. Ionic bonds are formed between cations and...
Solution: Resonance is the phenomenon that allows a molecule to be expressed in multiple ways, none of which fully explain the molecule's properties. The molecule's structure is called a resonance...
Solution: The positions of the atoms remain constant in canonical forms, but the positions of the electrons change. The positions of atoms change in the given canonical forms. As a result, they...
Solution: However, while the carbonate ion cannot be represented by a single structure, the properties of the ion can be described by two or more different resonance structures. The actual structure...
Solution: Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Solution: During the formation of a molecule, the extent of bonding that occurs between two atoms is represented by the bond strength of the molecule. As the bond strength increases, the bond...
Solution: Ammonia's central atom (N) has one lone pair and three bond pairs. In water, the central atom (O) has two lone pairs and two bond pairs. As a result, the two bond pairs repel the two lone...
Solution: $BeCl_2$ The central atom does not have a lone pair, but it does have two bond pairs. As a result, its shape is AB2, or linear. $BCl_3$ The central atom has three bond pairs but no lone...
Solution: Ionic bonds are formed when one or more electrons are transferred from one atom to another. As a result, the ability of neutral atoms to lose or gain electrons is required for the...
Solution: “Atoms can combine either by transferring valence electrons from one atom to another or by sharing their valence electrons in order to achieve the closest inert gas configuration by having...
Solution: The lewis dot structures are:
Solution: For S and S2- A sulphur atom has only 6 valence electrons, which is a very small number. As a result, the Lewis dot symbol for the letter S is The presence of a...
Solution: Nitrogen atoms have only five valence electrons in total. As a result, the Lewis dot symbol for N is Bromine, because the atom has only seven valence electrons. As a result,...
Solution: Boron atoms have only three valence electrons, which is a very small number. As a result, the Lewis dot symbols for B are as follows: The oxygen atom has only six valence...
Solution: Only two valence electrons exist in the magnesium atom. As a result, the Lewis dot symbols for Mg are as follows: Only one valence electron exists in the sodium atom. As a...
Answer: "A chemical bond is an attractive force that holds a chemical species' constituents together." For chemical bond formation, many theories have been proposed, including valence shell electron...
Solution:
Solution: (b) Group 14 components have 4 valence electrons. Thus, bunch oxidation status is $+4$. In any case, the lower oxidation state turns out to be progressively steady because of the inactive...
Solution: (b) Borax is a strong base salt (NaOH) and a feeble corrosive $\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)$. In this way, it is essential thing in nature.
Solution: Carbon dioxide CO2 can be ready in the lab through the activity of weaken hydrochloric corrosive on calcium carbonate. Their response is as per the following: CO2 is industrially ready by...
Solution: The balanced equations are as follow:
Solution: The salt given to litmus is antacid. $X$ is, subsequently, a salt with a solid base, and a feeble corrosive. When $X$ is warmed unnecessarily, it additionally enlarges to frame material...
Solution: The given metal $X$ gives sodium hydroxide to a white accelerate, and the encourage breaks up surpassing sodium hydroxide. $X$ must, consequently, be made of aluminum. The acquired white...
Solution: $\rightarrow$ CO $=$ Neutral $\rightarrow \mathrm{B}_{2} \mathrm{O}_{3}=$ Acidic Being acidic, it responds with bases to frame salts. It responds with $\mathrm{NaOH}$ to frame sodium...
Solution: Allotropy is the presence of a component in more than one structure, having diverse actual properties however similar substance properties. Diamond's solid 3-D construction makes it a...
Solution: A tomic sweep (in pm) Aluminum Gallium In spite of the fact that Ga has more than one shell than Al, it is more modest in size than Al. This is on the grounds that the $3 \mathrm{~d}$...
Solution: Carbon ionizing enthalpy (the primary component in bunch 14 ) is exceptionally high $(1086 \mathrm{~kJ}/\mathrm{mol})$. That is normal on account of its little size. Nonetheless, there is...
Solution: (a) Silicon is warmed with methyl chloride at high temperature within the sight of copper A class of organosilicon polymers called methyl-subbed chlorosilane $\mathrm{MeSiCl}_{3},...
Solution: (a) Borax is warmed firmly Borax goes through different changes when warmed. It is losing atoms and expands of water right away. Then, at that point, it turns into a clear fluid, which...
Solution: (a) Diborane $\mathrm{B}_{2} \mathrm{H}_{6}$ is a compound that does not have an electron. $\mathrm{B}_{2} \mathrm{H}_{6}$ just has 12 electrons $-6 \mathrm{e}^{-}$of $6 \mathrm{H}$...
Solution: Hydrogen fluoride is a covalent compound with an exceptionally solid intermolecular clinging to hydrogen. Along these lines, it doesn't give particles and doesn't break up aluminum...
Solution: The $B-C l$ bond is normally polar as a result of the distinction in the electronegativities of $\mathrm{Cl}$ and $B$. However the particle of $\mathrm{BCl}_{3}$ is non-polar. That is on...
Solution; In $\mathrm{BF}_{3}$, the length of the $\mathrm{B}-\mathrm{F}$ bond is more limited than that of the $\mathrm{B}-\mathrm{F}$ bond in $\mathrm{BF}_{4}^{-} \cdot \mathrm{BF}_{3}$ is an...
Solution: - Lead is an individual from bunch 14 of the occasional table. The two oxidation situations with bunch shows are $+2$ and $+4$. The $+2$ oxidation state turns out to be more steady when...
Solution: Diamond: 1. It has a glasslike grid 2. In precious stone, every carbon molecule is sp3 hybridized and attached to four other carbon particles through a sigma bond. 3. It has an unbending...
Solution: The condition of hybridization of carbon in: (a) $C O_{3}^{2-}$ c in $\mathrm{CO}_{3}^{2-}$ is sp $^{2}$ hybridized and is attached to 3 oxygen iotas. (b) Diamond Every precious stone...
Solution: For $C O_{3}^{2-}$ There are just 2 resounding designs for the bicarbonate particle. For $\mathrm{HCO}_{3}^{-}$
Solution: Amphoteric substances will be substances that display both acidic and essential characteristics. Since aluminum breaks up in the two acids and bases, it is said to have an amphoteric...
Solution: (I) $\mathbf{B} \boldsymbol{F}_{3}$ Boron will in general shape monomeric covalent halides in view of its little size and high electronegativity. These halides of boron generally have a...
Solution: After warming orthoboric corrosive at a temperature of $370 \mathrm{~K}$ or above, it is changed over into metaboric corrosive and, upon additional warming, yields boric oxide...
Solution: Boric corrosive is a powerless monobasic corrosive which acts as a Lewis corrosive. In this way, it's anything but a protic corrosive. $\mathrm{B}(\mathrm{OH})_{3}+2 \mathrm{HOH}...
Solution: Since it is a Lewis corrosive, $\mathrm{BCl}_{3}$ promptly goes through hydrolysis to frame boric corrosive. $\mathrm{BCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 3...
Solution: The electronic arrangement of boron is $n s^{2} n p^{1}$. It contains 3 electrons in its valence shell. In this way, it can frame just 3 covalent bonds which imply that there are just 6...
Solution: Thallium and boron have a place with bunch 13 of the intermittent table and $+1$ oxidation state turns out to be more steady as we drop down the gathering. Boron is more steady than...
Solution: (I) $B$ to $\mathrm{TI}$ Gathering 13 components have their electronic design of $\mathrm{ns}^{2} \mathrm{np}^{1}$ and the oxidation state displayed by these components ought to be 3 ....
Solution: The galvanic cell relating to the given redox response can be displayed as: $\mathrm{Zn}\left|Z n_{(a q)}^{2+} \| A g_{(a q)}^{+}\right| \mathrm{Ag}$ (I) Zn anode is contrarily charged on...
Solution: The diminishing specialist is more grounded as the terminal potential declines. Subsequently, the expanding request of the lessening force of the given metals is as given underneath: Ag...
Solution: A metal with more grounded diminishing force uproots one more metal with more vulnerable lessening power from its answer of salt. The request for the expanding diminishing force of the...
Solution: (I) In fluid arrangement, AgNO3 ionizes to give Ag+(aq) and NO3–(aq) particles. \[AgN03\left( aq \right)\text{ }\to \text{ }Ag+\left( aq \right)\text{ }+\text{ }NO3\left( aq...
Solution: (a) $F e_{(a q)}^{3+}$ and $I_{(a q)}^{-}$ $2 F e_{(a q)}^{3+}+2 I_{(a q)}^{-} \rightarrow 2 F e_{(a q)}^{2+}+I_{2(s)}$ Oxidation half response: $2 I_{(a q)}^{-} \rightarrow I_{2}(s)+2...
Solution: The reasonable response is as given underneath: $4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_{2} \mathrm{O}_{(g)}$ $4 N H_{3}=4 \times 17...
Solution: One of the responding components consistently has a component that can exist in somewhere around 3 oxidation numbers. (I) The non - metals which can show disproportionation responses are...
Solution: The redox response is as given beneath: Cl2(s)+SO2(aq)+H2O(l)→Cl(aq)-+SO4(aq)2- The oxidation half response: SO2(aq)→SO4(aq)2- Add 2 electrons to adjust the oxidation no. :...
Solution: The response is as given beneath: $\mathrm{Mn}_{(a q)}^{3+} \rightarrow \mathrm{Mn}_{(a q)}^{2+}+\mathrm{MnO}_{2(s)}+\mathrm{H}_{(a q)}^{+}$ The oxidation half response: $\mathrm{Mn}_{(a...
Solution: The oxidation no. of $\mathrm{C}$ in $(C N)_{2}, C N^{-}$and $C N O^{-}$are $+3,+2$ and $+4$ separately. Let the oxidation no. of $\mathrm{C}$ be $\mathrm{y}$. $(C N)_{2}$ $2(y-3)=0$ Along...
Solution:
Solution: (a) $M n O_{4}^{-}(a q)+I_{(a q)}^{-} \rightarrow \operatorname{MnO}_{2}(s)+I_{2}(s)$ Stage 1 The two half responses are given beneath: Oxidation half response: $I_{(a q)} \rightarrow...
Solution: $\mathrm{Ag}^{+}$and $C u^{2+}$ acts as oxidizing specialist in responses (I) and (ii) individually. In response (iii), $\mathrm{Ag}^{+}$oxidizes $\mathrm{C}_{6} \mathrm{H}_{5}...
Solution: $X e O_{6(a q)}^{4-}+2 F_{(a q)}^{-}+6 H_{(a q)}^{+} \rightarrow X e O_{3(g)}+F_{2(g)}+3 H_{2} O_{(l)}$ The oxidation no. of Xe decreases from $+8$ in $\mathrm{XeO}_{6}^{4-}$ to $+6$ in...
Solution: $F_{2}$ can oxidize $C l^{-}$to $C l_{2}, B r^{-}$to $B r_{2}$, and $I^{-}$to $I_{2}$ as: $F_{2(a q)}+2 C l_{(s)}^{-} \rightarrow 2 F_{(a q)}^{-}+C l_{2(g)}$ $F_{2}(a q)+2 B r_{(a q)}^{-}...
Solution: The normal oxidation no. of $\mathrm{S}$ in $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is $+2$. The normal oxidation no. of $\mathrm{S}$ in $S_{4} \mathrm{O}_{6}^{2-}$ is $+2.5$. The oxidation...
Solution: (a) $2 \mathrm{AgBr}_{(s)}+C_{6} H_{6} O_{2}(a q) \rightarrow 2 \mathrm{Ag}_{(s)}+2 \mathrm{HBr}_{(a q)}+C_{6} \mathrm{H}_{4} O_{2}(a q)$ $\mathrm{C}_{6} \mathrm{H}_{6}...
Solution: (a) While producing benzoic corrosive from toluene, alcoholic potassium permanganate is utilized as an oxidant because of the given reasons. (I) In an impartial medium, $O H^{-}$ions are...
Solution: When there is a response between lessening specialist and oxidizing specialist, a compound is framed which has lower oxidation number if the diminishing specialist is in abundance and a...
Solution: The oxidation no. of $A g$ in $A g F_{2}$ is $+2$. Be that as it may, $+2$ is entirely unsound oxidation no. of Ag. Consequently, when $A g F_{2}$ is framed, silver acknowledges an...
Solution: (a) Stage 1: $\mathrm{H}_{2} \mathrm{O}$ breaks to give $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$. $2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow 2...
Solution: In sulfur dioxide $\left(S O_{2}\right)$ the oxidation no. of $\mathrm{S}$ is $+4$ and the scope of oxidation no. of sulfur is from $+6$ to $-2$. Consequently, $S O_{2}$ can go about as a...
Solution: The compound where carbon has oxidation number from -4 to +4 are given below in the table:
Solution: Formulas are: (a) Mercury (II) chloride $H g C l_{2}$ (b) Nickel (II) sulphate $\mathrm{NiSO}_{4}$ (c) Tin (IV) oxide $\mathrm{SnO}_{2}$ (d) Thallium (I) sulphate $\mathrm{Tl}_{2}...
Solution: O.N. of S in H2SO5. By traditional strategy, the O.N. of S in H2SO5 is 2 (+1) + x + 5 (- 2) = 0 or x = +8 This is outlandish on the grounds that the most extreme O.N. of S can't be more...
Solution: $H_{2} O_{(s)}+F_{2}(g) \rightarrow H F_{(g)}+H O F_{(g)}$ In the above response, Oxidation no. of $\mathrm{H}$ and $\mathrm{O}$ in $\mathrm{H}_{2} \mathrm{O}$ is $+1$ and $-2$...
Solution: $4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_{2} \mathrm{O}_{(g)}$ In the above response, Oxidation no. of $\mathrm{N}$ and $\mathrm{H}$ in $N...
Solution: $2 K_{(s)}+F_{2(g)} \rightarrow 2 K+F_{(s)}$ In the above response, Oxidation no. of $\mathrm{K}$ is 0 . Oxidation no. of $F$ is 0 . Oxidation no. of $\mathrm{K}$ and $\mathrm{F}$ in...
Solution: $4 \mathrm{BCl}_{3(g)}+3 \mathrm{LiAlH}_{4(s)} \rightarrow 2 \mathrm{~B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}_{(s)}+3 \mathrm{AlCl}_{3}(s)$ the above response, Oxidation no. of...
Solution: $\mathrm{Fe}_{2} \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2}(g)$ In the above response, Oxidation no. of Fe and $\mathrm{O}$ in $\mathrm{Fe}_{2}...
Solution: $\mathrm{CuO}_{(s)}+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(g)}$ Oxidation no. of $\mathrm{Cu}$ and $\mathrm{O}$ in $C u O$ is $+2$ and $-2$...
Solution: $\underline{\mathrm{C}} \mathrm{H}_{3} \underline{\mathrm{C}} \mathrm{OOH}$ C2x+14O2 \mathrm{C}_{2}^{\mathrm{x}}+1_{4} \mathrm{O}_{2} Let $x$ be the oxidation no. of $C$. Oxidation...
Solution: $\mathrm{CH}_{3} \mathrm{C} \mathrm{H}_{2} \mathrm{OH}$ $$ $$ Let $x$ be the oxidation no. of $C$. Oxidation no. of $\mathrm{O}=-2$ Oxidation no. of $\mathrm{H}=+1$ Then, at that...
Solution: $\mathrm{Fe}_{3} \mathrm{O}_{4}$ Let $x$ be the oxidation no. of Fe. Oxidation no. of $0=-2$ Then, at that point, $3(x)+4(- 2)=0$ $3 x-8=0$ $x=\frac{8}{3}$ Oxidation no. can't be partial....
Solution: H2S4O6: Let x be the oxidation number of S. \[\begin{array}{*{35}{l}} Oxidation\text{ }no.\text{ }of\text{ }H\text{ }=\text{ }+1 \\ Oxidation\text{ }no.\text{ }of\text{ }O\text{ }=\text{...
Solution: KI3: In KI3, the oxidation number (O.N.) of K is +1. Consequently, the normal oxidation number of I is Ncert Solutions Cbse Class 11-science Chemistry Chapter - Redox Reactions. In any...
If the threshold wavelength is $\lambda_{0} n m\left(=\lambda_{0} \times 10^{-9} \mathrm{~m}\right)$, the K.E. of the radiation would be: $ h\left(\nu-\nu_{0}\right)=\frac{1}{2} m \nu^{2} $ Three...
Given, the work function of caesium $\left(W_{0}\right)=1.9 \mathrm{eV}$ (a)From the $W_{0}=\frac{h c}{\lambda_{0}}$ expression, we get: $\lambda_{0}=\frac{h c}{W_{0}}$ Where, $\lambda_{0}$ is the...
From the expression of energy of one photon (E), $ E=\frac{h c}{\lambda} $ Where, $\lambda$ denotes the wavelength of the radiation $\mathrm{h}$ is Planck's constant c denotes the velocity of the...
No. quanta in $2 \mathrm{~J}$ of energy $\frac{2 J}{32.27 \times 10^{-20} J}$ $ =6.19 \times 10^{18} $ $ =6.2 \times 10^{18} $
Energy of one quantum $(E)=h v$ $ =\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(4.87 \times 10^{14} \mathrm{~s}^{-1}\right) $ Energy of one quantum $(\mathrm{E})=32.27 \times 10^{-20}...
Speed of the radiation, $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ Distance travelled by the radiation in a timespan of $30 \mathrm{~s}$ $ \begin{array}{l} =\left(3 \times 10^{8}...
Wavelength of the emitted radiation $=616 \mathrm{~nm}=616 \times 10^{-9} \mathrm{~m}$ (Given) (a)Frequency of the emission $(\nu)$ $ \nu=\frac{c}{\lambda} $ Where, $c=$ speed of the radiation...
Energy with which it emits photons=Power of laser Power $=E=\frac{N h c}{\lambda}$ Where, $N=$ number of photons emitted $\mathrm{h}=$ Planck's constant $\mathrm{c}=$ velocity of radiation...
Let us consider the total no. electrons present in $A^{3+}$ be $x$. Now, total no. neutrons in it $=x+30.4 \%$ of $x=1.304 x$ Since the ion has a charge of $+3, \Rightarrow$ no. electrons in neutral...
The number of electrons in a negatively charged ion be $x$. Then, no. neutrons present $=x+11.1 \%$ of $x=x+0.111 x=1.111 x$ No. electrons present in the neutral atom $=(x-1)$ (When an ion carries...
Let us consider that the No.of protons in the element be $x$. No. of neutrons $=x+31.7 \%$ of $x$ $=x+0.317 x$ $=1.317 x$ According to the question, Mass number of the element $=81$, which...
The general convention followed while representing elements along with their atomic masses (A), and their atomic numbers $(Z)$ is ${ }_{Z}^{A} \mathrm{X}$. Therefore, ${ }_{35}^{79} \mathrm{Br}$ is...
The findings obtained with a foil made up of heavy atoms will differ from those obtained with a foil made up of comparatively light atoms. The magnitude of positive charge in the nucleus of a...
Charge held by the oil drop $=1.282 \times 10^{-18} C$ Charge held by one electron $=1.6022 \times 10^{-19} C$ Therefore, No. electrons present in the drop of oil $\frac{1.282 \times 10^{-18}...
Charge held by one electron $=1.6022 \times 10^{-19} C \Rightarrow 1.6022 \times 10^{-19} C$ charge is held by one electron. Therefore, electrons carrying charge of $2.5 \times 10^{-16} C \quad...
Radius of carbon atom$ =\frac{2.6}{2} $ $ \begin{array}{l} =1.3 \times 10^{-10} \mathrm{~m} \\ =130 \times 10^{-12} \mathrm{~m}=130 \mathrm{pm} \end{array} $ (b) Length of the arrangement $=1.6...
Length of the arrangement $=2.4 \mathrm{~cm}$ No. carbon atoms present $=2 \times 10^{8}$ The diameter of the carbon atom $=\frac{2.4 \times 10^{-2} m}{2 \times 10^{8} m}$ $=1.2 \times 10^{-10} m...
We know that$ 1 \mathrm{~cm}=10^{-2} \mathrm{~m} $ Length of the scale $=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$ Diameter of one carbon atom $=0.15 \mathrm{~nm}=0.15 \times 10^{-9}...
The energy associated with hydrogen-like species is: $ E_{n}=-2.18 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right) J $ For the ground state of the hydrogen atom, $ \begin{array}{l} \Delta...
The wave number associated with the Balmer transition for the He+ ion (n = 4 to n = 2 ) is given by: $ \bar{\nu}=\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) $...
Only one electron exists in hydrogen atoms. The angular momentum of this electron, according to Bohr's postulates, is: $m v r=n \frac{h}{2 \pi} \quad \ldots .(1)$ Where, $n=1,2,3, \ldots$ As per de...
(a)The total number of electrons in the atom =$2n^2$ if n is the primary quantum number. Hence, For n = 4, Total no. electrons = 2 (16) = 32 An atom with 32 electrons has the following electron...
(a)n = 1, l = 0 implies a 1s orbital. (b)n = 3 and l = 1 implies a 3p orbital. (c)n = 4 and l = 2 implies a 4d orbital. (d)n = 4 and l = 3 implies a 4f orbital.
(I) The range of potential values for 'l' is 0 to (n – 1). As a result, the possible values of l for n = 3 are 0, 1, and 2. The total number of potential ml = (2l + 1) values. It has a range of...
The number of electrons in H2 is 1 + 1 = 2. 2 – 1 = 1 number of electrons in H2+ H2: The number of electrons in H2 equals 1 + 1 = 2. Number of electrons in O2 = 8 + 8 = 16. In O2+, the number of...
(i)No.protons = no.electrons in a neutral atom. The number of protons in the atoms of the element is 29. (ii)The electronic configuration of this element (atomic number 29) is 1s...
For the 3d orbital: Principal quantum number ,possible values (n) = 3 Azimuthal quantum number, possible values(l) = 2 Magnetic quantum number ,possible values(ml) = – 2, – 1, 0, 1, 2
$(I I I)(a)[H e] 2 s^{1}$ Complete electronic configuration: $1 \mathrm{~s}^{2} 2 \mathrm{~s}{ }^{1}$. $\therefore$ the element's atomic number is 3 . The element is lithium (Li) (b) $[\mathrm{Ne}]...
The electronic configuration of the Hydrogen atom (in its ground state) $1 \mathrm{~s}^{1}$. The single negative charge on this atom indicates that it has gained an electron. Hence, the electronic...
As per de Broglie’s equation, $ \lambda=\frac{h}{m v} $ Given, K.E of electron $=3.0 \times 10^{-25} \mathrm{~J}$ $ \begin{array}{l} \text { Since K.E.= } \frac{1}{2} m v^{2} \therefore...
As per de Broglie’s equation, $ \lambda=\frac{h}{m v} $ Where, $\lambda$ denotes thr wavelength of the moving particle $\mathrm{m}$ is the mass of the particle $v$ denotes the velocity of the...
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as: $E_{5}=\frac{-\left(2.18 \times 10^{-18}\right)}{(5)^{2}}=\frac{-\left(2.18 \times 10^{-18}\right)}{25}=-8.72 \times...
Solution: For the union of balanced alkanes (for example alkanes with a significantly number of carbon particles), the Wurtz response is restricted. Two comparative alkyl halides are taken as...
Solution: The simplicity of nitration relies upon the presence of electron thickness on the compound to shape nitrates. Nitration responses are instances of electrophilic replacement responses where...
Solution: The response of benzene to the presence of Lewis acids (AlCl3) with an acyl halide or corrosive anhydride yields acyl benzene (or benzene ring). A Friedel-Craft alkylation response is...
Solution: Electrophiles are reagents that take part in a response by tolerating a couple of electrons to tie to nucleophiles. The higher the thickness of electrons on a benzene ring, the more...
Solution: Electrophiles are reagents that take part in a response by tolerating a couple of electrons to tie to nucleophiles. The higher the thickness of electrons on a benzene ring, the more...
Solution: The fundamental design of 2-methylbutane is displayed beneath: Based on the above structure, different alkenes that will give 2-methylbutane on hydrogenation are displayed beneath: a) b)...
Solution: Hexane into benzene: When vapours of hexane are passed over heated catalyst consisting of Cr₂O3 MO₂O3, and V₂O5, at 773 K under 10-20 atm pressure, cyclization and aromatization occurs...
Solution: Ethene into benzene: Ethene is first converted inti Ethyne and then into benzene.
Solution: Ethyne into benzene:
Solution: Fuel cell: Energy components are gadgets for creating power from fuel within the sight of an electrolyte. Dihydrogen can be utilized as a fuel in these cells. It is liked over different...
Solution: Water-gas shift response: It is a response of carbon monoxide of syngas blend with steam within the sight of an impetus as: $\mathrm{CO}_{(g)}+\mathrm{H}_{2} \mathrm{O}_{(g)} \rightarrow...
Solution: 'syngas': Syngas is a combination of carbon monoxide and dihydrogen. Since the combination of the two gases is utilized for the blend of methanol, it is called syngas amalgamation gas, or...
Solution: Hydrogenation: The method involved with adding dihydrogen to one more reactant is known as hydrogenation. It is utilized to decrease a compound within the sight of appropriate impetus. For...
Solution: Hydrogen economy: Dihydrogen delivers more energy than petroleum and is more eco-accommodating. Subsequently, it tends to be utilized in energy units to create electric force. Hydrogen...
Solution: Potassium chloride $(\mathrm{KCl})$ is the salt of a solid corrosive $(\mathrm{HCl})$ and solid base (KOH). Consequently, it is nonpartisan in nature and doesn't go through hydrolysis in...
Solution: The components of nuclear numbers 15 is phosphorus, 19 potassium, 23 is vanadium and 44 is ruthenium Hydride of Phosphorus Hydride of phosphorus $\left(\mathrm{PH}_{3}\right.$ ) is...
Solution: Normally, saline hydrides are ionic. Saline hydrides responds with water which brings about the arrangement of metal hydroxide alongside hydrogen gas freedom. It is addressed as, $A...
Solution: Hydration: The expansion of at least 1 atoms to a particle or particle which brings about arrangement of hydrated mixtures is known as hydration. For example $\mathrm{CuSO}_{4}+5...
Solution: Hydrogen peroxide goes about as an oxidizing specialist just as diminishing specialist in both basic medium and acidic medium. The response which are associated with oxidizing activities...
Solution: The amphoteric idea of water can be depicted based on the accompanying responses: 1) Reaction with $\mathrm{H}_{2} \mathrm{~S}$ The response happens as: $\mathrm{H}_{2}...
Solution: The method involved with treating extremely durable hardness of water utilizing engineered saps commonly based or trade of anions and cations present in water by $\mathrm{OH}^{-}$and...
Solution: (I) $\mathrm{PbS}_{(g)}+4 \mathrm{H}_{2} \mathrm{O}_{2(a q)} \rightarrow \mathrm{PbSO}_{4(s)}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}$ $\mathrm{H}_{2} \mathrm{O}_{2}$ is going about as an...
Solution: The response among water and fluorine can be addressed as: $2 F_{2(g)}+2 H_{2} O_{(l)} \rightarrow 4 H_{(a q)}^{+}+4 F_{(a q)}^{-}+O_{2(g)}$ This is an illustration of redox response $2...
Solution: Auto-protolysis (self-ionization) of water is a compound response wherein 2 water atoms respond to deliver a hydroxide particle $\left(\mathrm{OH}^{-}\right)$and a hydronium particle...
Solution: Saline hydrides [i.e.,LiH, NaH etc.] respond with water to shape hydrogen gas Furthermore, a base. The synthetic condition to address this response is $M H_{(g)}+H_{2} O_{(a q)}...
Solution: To shape an ordinary bond, an electron-lacking hydride doesn't have adequate electrons in which 2 electrons are shared by 2 particles. e.g., $B_{2} H_{6}$ $\mathrm{Al}_{2} \mathrm{H}_{6}$...
Solution: (i) $H_{2(g)}+M_{m} O_{0(g)} \rightarrow m M_{(s)}+H_{2} O_{(l)}$ (ii) $\mathrm{CO}_{(g)}+\mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{3} \mathrm{OH}_{(l)}$ (iii) $\mathrm{C}_{3}...
Solution: The readiness of di hydrogen is by the electrolysis of fermented or soluble water utilizing platinum anodes Generally, $15-20 \%$ of a corrosive $\left(\mathrm{H}_{2}...
Solution: 3-isotopes are: (i) tritium ${ }_{1}^{3} H$ or $\mathrm{T}$ (ii) protium ${ }_{1}^{3} H$ (iii) deuterium ${ }_{1}^{2} H$ or $\mathrm{D}$ Mass Ratio: Tritium : Protium : deuterium $=1: 2:...
Solution: Inductive Effect: The inductive impact alludes to the extremity delivered in a particle because of higher electronegativity of one molecule contrasted with another.Atoms or gatherings...
Solution: The bond cleavage can be displayed as: It goes under heterolytic cleavage since the common remaining parts with one of the sections. A carbocation is framed as the response middle.
Solution: The bond cleavage can be displayed as: It goes under heterolytic cleavage since the common remaining parts with the bromine particle. A carbocation is framed as the response...
Solution: The bond cleavage can be displayed as: It goes under heterolytic cleavage since the common remaining parts with the carbon molecule of propanone. A carbanion is framed as the response...
Solution: The bond cleavage can be displayed as: It goes under homolytic cleavage since one of the common pair in a covalent bond goes with the reinforced molecule. A free revolutionary is framed as...
Solution: The given mixtures are a pair of contributing constructions or authoritative structures. They don't address any genuine particle and are absolutely speculative. They are additionally...
Solution: The given mixtures are a pair of mathematical isomers since they have a similar molecular formulas, succession of covalent bonds and a similar constitution yet vary in the general...
Solution: The given mixtures are a pair of primary isomers since they have a similar molecular formula yet have various structures. These mixtures vary in the situation of the ketone bunch. For the...
Solution: \[\mathbf{C}{{\mathbf{H}}_{\mathbf{3}}}\mathbf{CH}=\mathbf{CHCH}2\] Resonating structures:
Solution: C6H5CH2 Resonating structures:
Solution: structure of C6H5CHO: Resonating structures:
Solution: \[C{{H}_{3}}CH\text{ }=\text{ }CH\text{ }\text{ }CHO\] Resonating structures:
Solution: structure of C6H5NO2: Resonating structures:
Solution: structure of C6H5OH : Resonating structures:
Solution: Because of hyperconjugation, an alkyl bunch acts as an electron-benefactor bunch when joined to a π framework. For instance, take a gander at propene. The sigma electrons of the C-H bond...
Solution: Nitro (–NO2), C=C Ethylenic Alkene
Solution: \[~Ester\text{ }-\left( O\text{ }=\text{ }C\text{ }\text{ }O \right)-,\text{ }{{1}^{o}}Amino\text{ }\left( N{{H}_{2}} \right)\text{ }\left( aromatic \right),\text{ }Diethylamine\text{...
Solution: Hydroxyl (–OH), Aldehyde (–CHO), Methoxy (–OMe), C=C double bond
Solution: Hexanedial Condensed Formula: \[\left( CHO \right){{\left( C{{H}_{2}} \right)}_{4}}\left( CHO \right)\] Bond line formula: Functional groups: Aldehyde (-CHO) group
Solution: 2–hydroxy–1, 2, 3–propanetricarboxylic acid Condensed Formula: \[\left( COOH \right)C{{H}_{2}}C\left( OH \right)\text{ }\left( COOH \right)C{{H}_{2}}\left( COOH \right)\] Bond line...
Solution: 2, 2, 4–trimethylpentane Condensed formula: \[{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{2}}C\text{ }{{\left( C{{H}_{3}} \right)}_{3}}\] Bond line formula : Functional group: no functional...
Solution: 2, 2–dichloroethanol
Solution: 3–chloropropanal
Solution: 3–bromo–3–chloroheptane
Solution: 2, 5–dimethyl heptane
Solution: 3–methylpentanenitrite
Solution: Propylbenzene
Solution: Bond line formula for Heptan–4–one:
Solution: Bond line formula for 2, 3–dimethyl butanal :
Solution: Bond line formula for Isopropyl alcohol:
Solution: HCONHCH3 There are four C–H sigma bonds, two C–N sigma bond, one N–H sigma bond, and one C=O pi bond in the given compound.
Solution: CH3NO2 There are three C–H sigma bonds, one C–N sigma bond, one N–O sigma bond, and one N=O pi bond in the given compound.
Solution: CH2 = C = CH2 There are two C–C sigma bonds, four C–H sigma bonds, and two C=C pi bonds in the given compound.
Solution: CH2Cl2 There are two C – Cl sigma bonds and two C–H sigma bonds in the given compound.
Solution: C6H12 There are six C – C sigma bonds and twelve C–H sigma bonds in the given compound.
Solution: C6H6 There are six C – C sigma bonds, six C–H sigma bonds, and three C=C pi resonating bonds in the given compound.
Let us consider f: A → B be many-one function. Let us consider f(a) = p and f(b) = p So, for inverse function we will have \[{{f}^{-1}}\left( p \right)\text{ }=\text{ }a\] and \[{{f}^{-1}}\left( p...
Answer: At equilibrium, the pressure exerted by ethane and hydrogen gas (each) is given by the symbol p. Now, according to the reaction, C2H6 (g) ↔ C2H4 (g) + H2 (g) Initial conc. 4.0atm ...
Answer: The equilibrium constant, Kc, is defined as the product of the equilibrium concentrations of products over the equilibrium concentrations of reactants, each raised to the power of the...
Answer: It is given that equilibrium constant $\mathrm{K}_{\mathrm{c}}$ for the reaction $\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{HI}_{(\mathrm{g})}$ is...
Answer: As we know that the equilibrium constant, Kc, is defined as the product of the equilibrium concentrations of products over the equilibrium concentrations of reactants, each raised to the...
Answer: For the given reaction, we have been given the information, $\Delta \mathrm{n}=2-3=-1$ $\mathrm{T}=450 \mathrm{~K}$ $\mathrm{R}=0.0831$ bar $\mathrm{L}$ bar $\mathrm{K}^{-1}...
Answer: The given reaction is: $2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{2(\mathrm{~g})}$ $\rightarrow$ $2 \mathrm{NOBr}_{(\mathrm{g})}$ Now, $2 \mathrm{~mol}$ of $\mathrm{NOBr}$ is formed...
Answer: Assuming the concentration of $\mathrm{N}_{2} \mathrm{O}$ at equilibrium be $\mathrm{x}$ The given reaction is: $2 \mathrm{~N}_{2}(\mathrm{~g})$+$\mathrm{O}_{2}(\mathrm{~g}) \quad...
Answer: The equilibrium constant statement does not include pure solids or pure liquids since they are inert. Because they have no effect on the quantity of reactant present at equilibrium in the...
Answer: Kp and Kc are equilibrium constants for reversible reactions. The equilibrium constant Kp is stated in terms of atmospheric pressure, whereas Kc is expressed in terms of concentrations...
Answer: The equilibrium constant, Kc, is defined as the product of the equilibrium concentrations of products over the equilibrium concentrations of reactants, each raised to the power of the...
Answer: The equilibrium constant, Kc, is defined as the product of the equilibrium concentrations of products over the equilibrium concentrations of reactants, each raised to the power of the...
Answer: Evaluating Partial pressure of lodine atoms (I) $p_{I}=\frac{40}{100} \times p_{\text {total }}$ $=\frac{40}{100} \times 10^{5}$ $=4 \times 10^{4} \mathrm{~Pa}$ Similarly, Partial pressure...
Answer: Given information in the question, $2 S O_{2}(g)+O_{2}(g) \rightleftharpoons 2 S O_{3}(g)$ As we know that, $K_{c}=\frac{\left[S O_{3}\right]^{2}}{\left[S...
Answer: Equality of rates of forward and backward processes restores equilibrium. However, the vapour pressure will stay unchanged because it is temperature dependent and not container volume....
Answer: (a) As the container's capacity increases, the vapor pressure decreases due to the increased distribution of vapors. (b) Increasing the container's capacity increases evaporation rates...