It is measured by potentiometer which does not have internal resistance.
\(\begin{array}{lr} 2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+6 \mathrm{e}^{-} \text {At anode } \\ { } \frac{3 \mathrm{Cu}^{2+}(a q)+6 e^{-} \longrightarrow 3 \mathrm{Cu}(s)}{2 \mathrm{Al}(s)+3 \mathrm{Cu}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \operatorname{Cu}(s)} \\ \hline \end{array}\)

Here, I^- is oxidised to I₂ and O₂(g) is reduced to OH^-.
\(4 \mathrm{I}+\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{OH}^{-}+2 \mathrm{I}_{2}\)

All the hybrid orbitals have same shape. However, their sizes are in the order : sp < sp² < sp³

The diamagnetic species are those which contain only;y paired electrons. This can be predicted from the molecular orbital electronic configuration as given below.
\( { H }_{ 2 }^{ + }:\sigma 1{ s }^{ 1 }\)
\(\mathrm{O}_{2}: \sigma^* \mathrm{ls}^{2}, \dot{\sigma}^* 1 s^{2}, \sigma^* 2 s^{2}, \dot{\sigma} ^*2 s^{2}, \sigma^* 2 p_{z}^{2}\)
\(\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}, \pi^{*} 2 p_{x}^{\approx} \pi 2 p_{x}^{1}, \pi^{*} 2 p_{x}=\pi^{*} 2 p_{y}^{1}\)
\(\mathrm{O}_{2}^{2+}: \sigma ^*1 s^{2}, \hat{\sigma} 1 s^{2}, \sigma 2 s^{2}, \hat{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}, \pi 2 p_{y}^{2}\)
Since, \({ O }_{ 2 }^{ 2+ }\) contain all paired electrons, it will be diamagnetic.

As we move from left to right in period, the number of valence electrons increases by one at each succeeding element but the number of shells remains same. Due to this, effective nuclear charge ivcreases.
More is the effective nuclear charge, more is the attraction between nuclei and electron.
Hence, the tendency of the element to lose electrons decreases, this results in decrease in metallic character.
Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non-metallic character increases on moving from left to right in a period.

No, the maximum number of elements which can be accommodated in the present set up of the long form of the periodic table is 118. Thereafter, filling of 8s-orbital shell begin which all accommodate only two electrons. After 8s-oorbitals, the filling of 5g-orbitals will begin. Since we do not have any provision for g-block elements in the present set up of the long form of periodic table, therefore, an element with atomic number 126, if discovered, cannot be accommodated in the present set up of the long form of periodic table.

v = 2.92 x 10¹⁵ s^-1;
\( R_H=109677 \mathrm{~cm}^{-1} \)
\( n_1=1, n_2=3 \)
\( \bar{v}=109677\left[\frac{1}{1^2}-\frac{1}{3^2}\right] \mathrm{cm}^{-1} \)
\(109677 \times \frac{8}{9} \mathrm{~cm}^{-1}=97490.7 \mathrm{~cm}^{-1} \)
\( \text { wavelength }(\lambda)=\frac{1}{\bar{v}}=\frac{1}{97490.7} \mathrm{~cm} \)
\( =1.03 \times 10^{-5} \mathrm{~cm}=1.03 \times 10^{-7} \mathrm{~m} \)
\( =103 \times 10^{-9} \mathrm{~m}=103 \mathrm{~nm} \)

(a) 1s, 2s and 3s - orbitals in Mg-atom are not degenerate because these have different values of n.
(b) 2p_x, 2p_y and 2p_z - orbitals in C-atom are degenerate because these belong to same subshell.
(c) 3s, 3p_x and 3d - orbitals in H - atom are degenerate because for H-atom, the subshells having same value of n have same energies.

(i) \(10 \mathrm{mg} \operatorname{sugar}\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)=0.01 / 342 \text { mole }=2.92  \times 10^{-5} \mathrm{~mole}\)
(ii) 0.05 cc water \(=0.05 \mathrm{~g}=0.05 / 18\) mole \(=2.78 \times 10^{-3}\) mole

[Ne]3s²3p³, it is due to smaller size and stable electronic configuration.

Be (4) has electronic configuration 1s² 2s² whereas B(5) has 1s² 2s² 2p¹.
The energy required to remove electron from completely filled 2s orbital is higher than the energy required to remove electron from 2p orbital.

n = 5 1 = 0, 1, 2, 3, [∴ 1 = 0, 1, .... n - 1 but I = 4 is not possible 'g' orbital is not being orbital used]
n = 5, I = 0 5s n = 5, I = 1 is 5p.
n = 5, 1 = 2 is 5d n = 5, 1 = 3 is 5f

EMF of a cell is the difference in the electrode potentials of the two electrodes in a cell when no current flows through the cell.

₈O = 1 s² 2s²  2p⁴

1. Molecular mass of carbon dioxide = 44 g
2. Therefore, the mass of 1 mol of CO₂ is 44 g. As per Avogadro's rule, 1 mol of substance has \(6.022 \times 10^{23}\) molecules.
3. Therefore mass of \(6.022 \times 10^{23}\) molecules of CO₂ is 44 g.
4. Hence, the mass of 1 molecule of CO₂ is \(=\frac{44}{6.022 \times 10^{23}}=7.306 \times 10^{-23} \mathrm{~g}\)

According to molecular orbital theory, electronic configurations and bond order of \(\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}, \mathrm{O}_{2}\) and \(\mathrm{O}_{2}^{+}\)species are as follows. 
\(\mathrm{N}_{2}\left(14 e^{-}\right)=\sigma 1 s^{2}, \stackrel{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \stackrel{\star}{\sigma} 2 s^{2}\) 
\(\left(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}\right), \sigma 2 p_{z}^{2}\)
Bond order = \(\frac{1}{2}\left[N_{b}-N_{a}\right]=\frac{1}{2}(10-4)=3\)
\(\mathrm{N}_{2}^{+}\left(13 e^{-}\right)=\sigma 1 s^{2}, \stackrel{*}{\sigma 1 s}^{2}, \sigma 2 s^{2}, \stackrel{*}{\sigma} 2 s^{2},\left(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}\right) \sigma 2 p_{z}^{1}\)
\(\text { Bond order }=\frac{1}{2}\left[N_{b}-N_{a}\right]=\frac{1}{2}(9-4)=2.5\)
\(\mathrm{O}_{2}\left(16 \mathrm{e}^{-}\right)=\sigma 1 s^{2},{ }^{*} 1 s^{2}, \sigma 2 s^{2}, \stackrel{*}{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}\)
\(\left(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}\right),\left(\pi 2 p_{x}^{1} \approx \pi 2 p_{y}^{1}\right)\)
\(\text { Bond order }=\frac{1}{2}\left[N_{b}-N_{a}\right]=\frac{1}{2}(10-6)=2\)
\(\mathrm{O}_{2}^{+}\left(15 e^{-}\right)=\sigma 1 s^{2}, \sigma 1 s^{2}, \sigma 2 s^{2}, \sigma 2 s^{2}, \sigma 2 p_{z}^{2}\)
\(\left(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}\right),\left(\pi 2 p_{x}^{1} \approx \pi 2 p_{y}\right)\)
\(\text { Bond order }=\frac{1}{2}\left[N_{b}-N_{a}\right]=\frac{1}{2}(10-5)=2.5\)
(i) \(\begin{array}{c} \mathrm{N}_{2} \longrightarrow \mathrm{N}_{2}^{+}+e^{-} \\ \mathrm{BO}=3 \quad \mathrm{BO}=2.5 \end{array}\)
Thus, bond order decreases.
(ii) \(\underset{\mathrm{BO}=2}{\mathrm{O}_{2}} \longrightarrow \underset{\mathrm{BO}=2.5}{\mathrm{O}_{2}^{+}}+e^{-}\)
Thus, bond order increases

The ionisation enthalpies decreases regularly as we move down a group from one element to the other. This is evident from the values of the first ionisation enthalpies of the elements of group 1 (alkali metals) and group 17 elements as given in table and figure.

Given trend can be easily explainedon the basis of increasing atomic size and screening effect that are as follows.
(i) On moving down the group, the atomic size increases gradually due to the addition of one new principal energy shell at each succeeding element. Hence, the distance of the valence electrons from the nucleus increases. Consequently, the force of attraction between the nucleus valence electrons decreases and hence, the ionisation enthalpy decreases.
(ii) With the addition of new shells, the shielding or the screening effect increases. As a result, the force of attraction between the nucleus and the valence electrons further decreases and hence, the ionisation enthalpy decreases.
(iii) Nuclear charge increases with increase in atomic number. As a result, the force of attraction by the nucleus for the valence electrons should increase and accordingly the ionisation enthalpy should increase. The combined effect of the increase in the atomic size and the screening effect more than compensates the effect of the increased nuclear charges. Consequently, the valence electrons become less and less firmly held by the nucleus and hence, the ionisation enthalpies gradually decreases as move down the group.

(a) No. of σ bonds = 7
No. of \(\pi\) bonds = 3
(b) Both the ions have the same bond order (0.5) but they differ in their configuration.
\({ H }_{ 2 }^{ + }\) ion = [ σls]¹,
\({ H }_{ 2 }^{ - }\) ion = [ σ1s]² [ σ x = s]¹
Since, \({ H }_{ 2 }^{ - }\) ion has an electron in the antibonding molecular orbital, it is therefore less stable.
The molecular orbital configuration of B₂ is given
B₂: [σ1s]² [σ*1s]² [σ2s]² [σ*2s]² [\(\pi\)2px]¹[\(\pi\) x 2py]¹
Since, B₂ has two unpaired electrons, it is paramagnetic

(a) Writing the On on each atom above its symbol, then
\(3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ H } \overset { +5 }{ N } \overset { -2 }{ { O }_{ 3 } } (aq)\longrightarrow \overset { 0 }{ { Cl }_{ 2 }(g)+ } \overset { +3 }{ N } \overset { -2 }{ O } \overset { -1 }{ Cl(g)+ } \overset { +1 }{ { 2H }_{ 2 } } \overset { -2 }{ O } (l)\)
Here, the On of Cl increases from -1 in HCl to O in Cl₂ , therefore, Cl^- is oxidised and hence, HCl acts as the reducing agent. The ON of N decreases from +5 in HNO₃ to +3 in NOCL, therefore, HNO₃ acts as the oxidising agent. Thus this reaction is a redox reaction.
(b) Writing the ON of each atom above its symbol, we have,
\(\overset { +2 }{ Hg } \overset { -1 }{ { Cl }_{ 2 } } (aq)+\overset { +1 }{ 2K } \overset { -1 }{ I } (aq)\longrightarrow \overset { +2 }{ Hg } { \overset { -1 }{ I } }_{ 2 }(s)+2\overset { +1 }{ K } \overset { -1 }{ { Cl }^{ - }(aq) } \)
Here, the On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(c) \(\overset { +3 }{ { Fe }_{ 2 } } { \overset { -2 }{ O } }_{ 3 }(s)+3\overset { +2 }{ C } \overset { -2 }{ O } (g)\ \overset { \Delta }{ \longrightarrow } \ 2\overset { 0 }{ Fe } (s)+3\overset { +4 }{ C } \overset { -2 }{ { O }_{ 2 } } (g)\)
Here, On of decreases from +3 in Fe₂O₃ to 0 in Fe, therefore, Fe₂O₃ acts as an oxidising agent. Further, On of C increases from +2 in CO to +4 in CO₂, therefore, CO acts as a reducing agent. Thus, this reaction is an example of redox reaction.
(d) Writing the ON of each atom above its symbol, then
\(\overset { +3 }{ P } { \overset { -1 }{ Cl } }_{ 3 }(l)+3\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\quad \longrightarrow \quad 3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ { H }_{ 3 } } \overset { +3 }{ P } \overset { -2 }{ { O }_{ 3 } } (aq)\)
Here, On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(e) Writing the ON of each atom above its symbol, then
\(4\overset { -3 }{ N } \overset { +1 }{ { H }_{ 3 } } (aq)+3\overset { 0 }{ { O }_{ 2 } } (g)\quad \longrightarrow \quad 2\overset { 0 }{ { N }_{ 2 } } (g)+6\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\)

(i)  Ni = + 2, Pt = + 4
(II) Platinum forms more stable complexes in +4 state due to its higher stability than +2 state.

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

(a) The cell in which chemical energy of redox reaction is converted into electrical energy.
(b) It is because loss of electrons take place at anode, i.e., oxidation takes place.
(c) Lead storage battery
(d) It is because these metals are good reducing agents, cannot be obtained by chemical reduction.
(e) Electrons flow from anode to cathode where as current flows from cathode to anode.
(f) When concentration of each species is unity, any gas involved is at 1 bar, temperature is 298 K, the potential of electrode is called standard electrode potential measured with respect to standard hydrogen electrode.
(g) It means redox couple is stronger reducing agent than H⁺/H₂ couple.
E.g. \(\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=\) -0.76 V, Zn is stronger reducing agent than H₂.

(a) Number of orbitals = n ² = 3 ² = 9 orbitals, 3s, 3 p_x, 3 p_y , 3 p_z, 3 d _{x² - y²,} 3 dz², 3 dxy, 3 dyz and 3 dzx.
(b) (i) 2p (ii) 4s
(c) Orbital can have maximum two electron which must be of opposite spin.
(d) 's' orbitals are spherical and 'p' orbitals have dumb-bell shaped.
(e) d _{x² - y²} , dz².