NEET Chemistry Solved Question Paper 2013

Total number of antibonding electrons present in O₂ is 6.

$\left(\mathrm{\sigma }1\mathrm{s}{)}^2\right(\mathrm{\sigma }*1\mathrm{s}{)}^2\left(\mathrm{\sigma }2\mathrm{s}{)}^2\right(\mathrm{\sigma }*2\mathrm{s}{)}^2\left(\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}{)}^2\right(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\left)\right(\mathrm{\pi }*2{{\mathrm{p}}_{\mathrm{x}}}^1 = \mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}}^1)$

Maximum number of unpaired electrons are present in Gd³⁺. It has 7 unpaired electrons.

Yb²⁺ has no unpaired electrons.

Tb²⁺ has 5 unpaired electrons.

Pm³⁺ has 4 unpaired electrons.

La(OH)₃ is the most basic as it has the largest atomic size. Al(OH)₃ is amphoteric in nature.

K₂Cr₂O₇ in acidic medium converts into Cr³⁺. Since it is an oxidising agent, it undergoes reduction, i.e. decrease in oxidation state.

Cr₂O${}_7^{2-}$ + 14H⁺ + 6e^- $\to$2Cr³⁺ + 7H₂O

In BF₃, there is back bonding in between fluorine and boron due to presence of p-orbital in boron. Back bonding imparts double bond characteristics.

As BF, forms adduct the back bonding is no longer present and thus double bond characteristic disappears. Hence, bond becomes a bit longer than earlier (1.30 Å).

Bond dissociation energy is F₂ < Cl₂ because of relatively large electron-electron repulsion among the lone pairs in F₂ molecule where they are much closer to each other
than in Cl₂.

AX₃ $\rightleftharpoons$ $\underset{\mathrm{S}}{{\mathrm{A}}^{3+}} + \underset{3\mathrm{S}}{3{\mathrm{X}}^{-}}$

K_sp = [A³⁺] [X^-]³

      = (S).(3S)³ = 27S⁴

Given : $\frac{{\mathrm{\lambda }}_{\mathrm{particle}}}{{\mathrm{\lambda }}_{\mathrm{electron}}}$ = 1.8 $\times$ 10^-4

and $\frac{{\mathrm{\nu }}_{\mathrm{particle}}}{{\mathrm{\nu }}_{\mathrm{electron}}}$ = 3

According to de-Brogile equation, $\mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{m\nu }}$

$$\begin{gathered} \frac{{\mathrm{\lambda }}_{\mathrm{particle}}}{{\mathrm{\lambda }}_{\mathrm{electron}}} = \frac{\mathrm{h}}{{\mathrm{m}}_{\mathrm{particle}} \times {\mathrm{\nu }}_{\mathrm{particle}}} \times \frac{{\mathrm{m}}_{\mathrm{electron}} \times {\mathrm{\nu }}_{\mathrm{electron}}}{\mathrm{h}} \\ = \frac{{\mathrm{m}}_{\mathrm{electron}}}{{\mathrm{m}}_{\mathrm{particle}}} \times \frac{{\mathrm{\nu }}_{\mathrm{electron}}}{{\mathrm{\nu }}_{\mathrm{particle}}} \end{gathered}$$

$$\begin{gathered} \Rightarrow 1.8 \times {10}^{-4} = \frac{9.1 \times {10}^{-31 }\mathrm{kg}}{{\mathrm{m}}_{\mathrm{particle}}} \times \frac{1}{3} \\ {\mathrm{m}}_{\mathrm{particle} }= \frac{9.1 \times {10}^{-31}}{1.8 \times {10}^{-4} \times 3} = 1.6852 \times {10}^{-27} \mathrm{kg} \end{gathered}$$

Actual mass of neutron is 1.67493 $\times$ 10^-27 kg. Hence, the particle is neutron.

H  and U are state functions but W and q are not state functions.

From the equation, $∆\mathrm{H} = ∆\mathrm{U} + ∆\mathrm{PV}$

At constant pressure, $∆\mathrm{H} = ∆\mathrm{U} + \mathrm{P}∆\mathrm{V}$

At constant volume, $∆\mathrm{H} = ∆\mathrm{U} + \mathrm{V}∆\mathrm{P}$

At constant pressure, $∆\mathrm{P} = 0, ∆\mathrm{H} = {\mathrm{q}}_{\mathrm{p}}$

so, it is a state function.

At constant volume, $∆$V = 0, $∆$U = q_v

so, it is a state function.

Work done in any adiabatic process is state function.

$∆$U = q - W ( $\because$ q = 0)

$∆$U = -W

Work done in isothermal process is not a state function.

W = -q ($\because$ $∆$T = 0, q $\ne$ 0)

Cu has lowest melting point because it has lowest enthalpy of atomisation (i.e., heat required to break the metal lattice to get free atoms) among the elements.