Electronic configuration of Be = 1s² 2s²
M.O. configuration of Be₂ = \((\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\)\( \left(\sigma^{*} 2 s\right)^{2} \)
\(\text { Bond order of } \mathrm{Be}_{2}=\frac{1}{2}(4-4)=0 \)
Since bond order is zero it does not exist.

(i) Group 17 elements (ns²np⁵) because they can gain electron easily.
(ii) Group 1 element (ns¹) because they can lose electron easily.
(iii) Group 18 elements have stable electronic configuration. They cannot lose or gain electron.

(a) The oxidation number of nitrogen in HNO₃ is +5 thus increase in oxidation number +5 does not occur hence HNO₃ cannot act as reducing agent but acts as an oxidising agent. In HNO₂ oxidation number of nitrogen is +3, it can decrease or increase with range of -3 to +5, hence it can act as both oxidising and reducing agent.
(b) Chlorine is in maximum oxidation state +7 in CIO₄ so it does not show the disproportionation reaction.
(c) Because it decomposes to give nascent oxygen.

(i) In third period, 3s and 3p orbitals are filled.
(ii) Promethium (Pm) with atomic number 61 is a man-made lanthanoid.
(iii) Actinoid series (f-block elements).

For the first shell \(l=2,{ m }_{ 1 }=-2,-1,0,+1,+2,{ m }_{ s }=\pm { 1 }/{ 2 }.-{ 1 }/{ 2 }\).it can have 2 electrons both with opposite spins.
\(For\quad n=2,\)
\( l=0,{ m }_{ l }=0,{ m }_{ s }=+{ 1 }/{ 2 },{ -1 }/{ 2 }\)
\( l=0,{ m }_{ 1 }=-1,0,+1,{ m }_{ s }=+{ 1 }/{ 2 },{ -1 }/{ 2 }.\)
\(Therefore, \ a \ total \ of \ 2+6=8 \ electrons \ are \ present\)
\(For\quad n=3,\quad when\)
\( l=0,{ m }_{ l }=0,{ m }_{ s }=+{ 1 }/{ 2 },{ -1 }/{ 2 }\)
\( l=0,{ m }_{ 1 }=-1,0,+1,{ m }_{ s }=+{ 1 }/{ 2 },{ -1 }/{ 2 }.\)
\( l=2,{ m }_{ 1 }=-2,-1,0,+1,+2,{ m }_{ s }=\pm { 1 }/{ 2 }.-{ 1 }/{ 2 }\)
\( Therefore,a \ total \ of \ 2+6+10=18 \ electrons \ are \ present.\)
\( For \ n=4,when\)
\(l=0,{ m }_{ 1 }=-1,0,+1,{ m }_{ s }=+{ 1 }/{ 2 },{ -1 }/{ 2 }.\)
\( l=0,{ m }_{ 1 }=-1,0,+1,{ m }_{ s }=+{ 1 }/{ 2 },{ -1 }/{ 2 }.\)
\(l=2,{ m }_{ 1 }=-2,-1,0,+1,+2,{ m }_{ s }=\pm { 1 }/{ 2 }.-{ 1 }/{ 2 }\)
\(l=3,{ m }_{ 1 }=-3,-2,-1,0,+1+2,+3,\)
\( { m }_{ s }=\pm { 1 }/{ 2,-{ 1 }/{ 2 } }\)
\(Therefore \ a \ total \ of \ 2+6+10+14=32 \ electrons \ are \ present.\)

(i) Energy of quantum (or photon),
E = hv = 6.626 x10^-34 J s x 4.870 x 10¹⁴_s^-1
= 32.268 x 10^-20J
(ii) Number of quanta present = \(\frac { Total\ energy\ produced }{ Energy\ of\ 1\ quanta } \)
= \(\frac { 2J }{ 32.27\times10^{ -20 }J } \simeq \ 6.2x10^{ 18 }quanta\)

Formula mass of \(\mathrm{NaNO}_3=23+14+48=85\)
\( 0.38 \mathrm{gNaNO}=23+14+48=85 \)
\( 0.38 \mathrm{gNaNO}=\frac{0.38}{85} \mathrm{~mol}=0.0045 \mathrm{~mol} \)
\( \text { Molarity of the solution }=\frac{0.0045 \mathrm{~mol}}{0.050 \mathrm{~L}}=0.090 \mathrm{M} \)

It is due to planar structure, individual dipoles get canceled.

Since, in the above reaction, the same element undergoes oxidation as well as reduction, so it is an example of disproportionation reaction.

This is due to the presence of dust particles.

Helium nucleus contains 2 protons and charge of a proton is same as that of an electron.
Therefore, the charge on the nucleus of a helium atom is (+2)×4.8×10⁻¹⁰=+9.6×10⁻¹⁰ esu

The equation for the given reaction is
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{FeSO}_{4}+\underset{\begin{array}{l} \text { Shows acidic } \\ \text { medium } \end{array}}{\longrightarrow}\) Cr³⁺ +Fe₂(SO₄)₃
Balancing of this equation by ion electron method involves the following steps.
Step I Converting unbalanced equation into ionic form It can be achieved by splitting the ionic compounds into constituent ions and by cancelling spectator ions.
\( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \)
Here \(\mathrm{SO}_{4}^{2-}\) is the spectator ion so after leaving it, the reaction becomes

Step II Splitting the equation into two halves

Step III Balancing all elements other than H and O
Here, oxidation half reaction is already balanced with respect to Fe atoms but to balance reduction half reaction, we have to multiply the Cr³⁺ by 2.
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)\)
Step IV Balancing H and O-atoms In oxidation half reaction, there is no need to balance H and a atoms, as in it these are not present. In reduction half reaction, we have to add 7H₂O towards product side to balance a-atoms and then add 14H⁺ towards reactant side to balance H-atoms.
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\underset{\underset{\mathrm{o}}{\mathrm{H}} \text { balance }}{14 \mathrm{H}^{+}} \longrightarrow 2 \mathrm{Cr}^{3+}+\underset{\begin{array}{l} \text { to balance } \\ \text { O-atoms } \end{array}}{7 \mathrm{H}_{2} \mathrm{O}}\)
Step V Balancing of charge
(i) \(\underset{+2}{\mathrm{Fe}}^{2+} \longrightarrow \underset{+3}{\mathrm{Fe}^{3+}}\)
So, we have to add one electron towards Fe³⁺ side.
Thus, \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}\)

So, we have to add 6e^- towards the reactant side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)
(ii) \(\left[\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}\right] \times 6 \text { to }\) equalise number of electrons.
Step VI Adding the half reactions On adding the half reaction, we get the overall balanced equation.
\( 6 \mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow 6 \mathrm{Fe}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l) \)
Step VII Verification of balanced equation Since, the number of atoms of each element are equal on both the sides, so it is a balanced chemical equation.
Further; \( 6 \mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow \mathrm 6 {Fe}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l) \)
6 x ( + 2)+ (- 2) + 14 (+ 1)\(\equiv\) 6 x ( + 3) + 2 x ( + 3) 12 - 2 + 14 \(\equiv\) 18 + 6 \(\Rightarrow\) 24 = 24
Thus, the equation is also balanced with respect to charge.
Step VIII Conversion of ionic equation into molecular form As we know \(\mathrm{SO}_{4}^{2-} \) is the spectator ion, and there are 6 \(\mathrm{Fe}^{2+}\) (Which require 6 \(\mathrm{SO}_{4}^{2-} \) ions) and 14H⁺ions (that require 7 \(\mathrm{SO}_{4}^{2-} \)ions), so, we have to add 13 \(\mathrm{SO}_{4}^{2-} \) ions on both the sides.
\(6 \mathrm{Fe}^{2+}(a q)+6 \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q)\) \( =+7 \mathrm{SO}_{4}^{2-}(a q) \longrightarrow 6 \mathrm{Fe}^{3+}(a q)+9 \mathrm{SO}_{4}^{2-}(a q) +2 \mathrm{Cr}^{3+}+3 \mathrm{SO}_{4}^{2-}(a q)+7 \mathrm{H}_{2} \mathrm{O}+\mathrm{SO}_{4}^{2-} \)
\( \text { or } 6 \mathrm{FeSO}_{4}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+7 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+7 \mathrm{H}_{2} \mathrm{O}+\mathrm{SO}_{4}^{2-} \)

(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

Writing the oxidation number of each element above its symbol in the following reactions
\(\overset { +2-2 }{ \underset { Basic \ oxide }{ 2PbO } } +\overset { +1-1 }{ \underset { Acid }{ 4HCl } } \longrightarrow \overset { +2-1 }{ 2Pb{ Cl }_{ 2 } } +\overset { +1 }{ { 2H }_{ 2 } } \overset { -2 }{ O } \)
In this reaction, oxidation number of each element remains same hence, it is not a redox reaction. In fact, it is an example of acid-base reaction .
\(\overset { +4 }{ Pb } \overset { -2 }{ { O }_{ 2 } } +\overset { +1-1 }{ 4HCl } \longrightarrow \overset { +2 }{ Pb } \overset { -1 }{ { Cl }_{ 2 } } +\overset { 0 }{ { Cl }_{ 2 } } +\overset { +1 }{ { 2H }_{ 2 } } \overset { -2 }{ O } \)
In PbO₂ Pb is in +4 oxidation state than +4 Pb in +2 oxidation state is more stable. So, Pb in +4 oxidation state (PbO₂ ) acts as an oxidising agent.
It oxidises Cl^- and Cl₂ and itself gets reduced to Pb²⁺

(i) Shape of \({ NH }_{ 2 }^{ - }\)
Number of valence electrons on central N atom = 5 + 1 (due to one unit negative charge) 
Number of atoms linked to it = 2
Total number of electron pairs around
\(N=\frac { 6+2 }{ 2 } =4\) and number of bond pairs = 2
Number of lone pairs = 4 - 2 = 2. Thus, the ion is of the type AB₂E₂. Hence, it has bent shape (V-shape).​​​​​​
(ii) Shape of O₃
While predicting geometry of molecules containing the double (or multiple) bond is considered as one electron pair. e.g. in case of ozone, its two resonating structures are

Thus, the central O-atom is considered to have two bond pairs and one lone pair, i.e. it is of the type AB₂ E. Hence, it is a  bent molecule. Thus, the two reasonating structures will be

1. D) A and C
2. C) International units of measurement
3. B) Chemical methods
4. C) Average mass of isotopes based on natural abundance
5. B) It must be balanced according to the laws of chemical combination

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(d) Assertion is wrong statement but Reason is correct statement.

(a) It is because \((\mathrm{N} \equiv \mathrm{N})\) , N2 has triple bond which has higher bond dissociation enthalpy than \(\mathrm{O}_{2}(\mathrm{O}=\mathrm{O})\) which has double bond.
(b) It is because its bond order is zero.
\(\text { B.O. }=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-10)=0\) 
(c) It is because F₂ does not have unpaired electron whereas O₂ has unpaired electron.
(d) F₂< B₂ < O₂ < C₂ < N₂. Higher the bond order, more is stability, more is bond dissociation enthalpy.
(e) It is because F₂ has lower bond dissociation enthalpy than O₂.
(f) 

(g) The double bond in C₂ consist of both \(\pi\) bonds because of the presence of four electrons in two \(\pi\) molecular orbitals.
(e) C₂ and Li₂ molecules are detected only in vapour phase.

(a) It is linear molecule, dipoles are equal and opposite, net dipole moment is zero.\(\mathrm{O} \stackrel{\leftrightarrow}{=} \mathrm{C} \stackrel{\leftrightarrow}{=} 0\)
(b) In SF₆,'S' has sp³d² hybridisation, octahedral shape.

(c) It is due to resonance.
(d) It is because o-nitrophenol has weak intra molecular H-bonds, where as p-nitrophenol has little stronger inter molecular H-bonds.
(e) 'S' is bigger in size, less electronegative than O.
(f) In \(\sigma\) bond extent of overlapping is more than \(\pi\)-bond.
(g) NaI < NaBr < NaCl < NaF.

(a) It is because they have large size, least effective nuclear charge and electropositive elements.
(b) It is due to inter electronic repulsion in 'O' due to exceptionally small size than 'S', therefore, less energy is released on gaining electron.
(c) It is because they have stable electronic configuration and energy is needed to overcome repulsion between valence electrons and electron to be added.
(d) P < S < F < Cl.

(a) 1 mole of N₂ needs 3 moles of H₂ ,
Therefore, 2 moles of N₂ needs 6 moles of H₂ .
But we have only 2 moles of H₂ , so H₂ is limiting reagent.
3 moles of H₂ gives 2 moles of NH₂.
2 moles of H₂ gives \(\frac{2}{3} \times 2=1.33\) moles of NH₃ .
(b) 18 g of water (1 mole) containing 10 x 6.022 x 10²³ electrons = 6.022 x 10²⁴ electrons.
[ \(\because\)1 molecule of H₂O = 2 + 8 = 10 e)
(c) 'M is molarity
'd is density of solution
'm' is molality
\(m=\frac{M \times 1000}{1000 \times d-M \times \text { Molar mass }}\)
\(=\frac{1 \times 1000}{1000 \times 1.10-1 \times 58.5}\)
\(=\frac{1000}{1100-58.5}=\frac{1000}{1041.5}=m=0.96 \mathrm{~mol} / \mathrm{kg} .\)
(d) It is ratio of number of moles of solute (component) to the total number of moles of solute and solvent (all components).
(e) X_glucose + x_H2O = 1
\(\Rightarrow\) X_H2O = 1 - 0.1 = 0.9