2 Marks Questions

Question 12 Marks

Write the electronic configuration of ${ }_9 F^{19},{ }_{16} S^{32}$ and ${ }_{18} Ar ^{38}$ and then point out the element with
i. maximum nuclear charge.
ii. minimum number of neutrons.
iii. maximum number of unpaired electrons.

Answer

$
\begin{aligned}
& { }_9 F^{19}=1 s^2 2 s^2 2 p_x^2 2 p_y^2 2 p_z^1, \\
& { }_{16} S^{32}=1 s^2 2 s^2 2 p^6 3 s^2 3 p_x^2 3 p_y^1 3 p_z^1, \\
& { }_{18} Ar^{38}=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6
\end{aligned}
$
i. Maximum nuclear charge $=18$ in ${ }_{18} Ar ^{38}$.
ii. Minimum number of neutrons $=10$ in ${ }_9 F^{19}$.
iii. Maximum number of unpaired electrons $=2$ in ${ }_{16} S^{32}$.

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Question 22 Marks

How would you convert cyclohexane to benzene?

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Question 32 Marks

What do you understated by Resonance energy?

Answer

Resonance energy: The difference between the energy of the most stable contributing structure and the energy of the resonance hybrid is known as resonance energy.
Example: The resonance energy of benzene is $147 KJ / mole$.

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Question 42 Marks

What does the following prefixes stand for -
a. pico
b. nano
c. centi
d. deci

Answer

S.No.	Prefix	Value
1.	Pico	$10^{-12}$
2.	nano	$10^{-9}$
3.	centi	$10^{-2}$
4.	deci	$10^{-1}$

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Question 52 Marks

Why are elements at the extreme left and extreme right the most reactive?

Answer

Elements at the extreme left and extreme right of periodic table are most reactive. It is due to following reasons:
1. Alkali metals are present at extreme left end of periodic table. Due to greatest atomic size of alkali metals, their ionisation enthalpy is low. These metals can easily lose electrons to form cations. Hence, these metals are very reactive in nature.
2. On the other hand, Halogens (group $17^{\text {th }}$ ) are present on extreme right side of periodic table (Noble gases, group $18^{\text {th }}$, are stable). Due to smallest atomic size of halogens, their electron gain enthalpy is very high. Therefore, these elements can form anions easily by gaining the electrons. So, These elements are also reactive.

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Question 62 Marks

For the equilibrium,
$
2 NOCl(g) \rightleftharpoons 2 NO(g)+Cl_2(g)
$
the value of the equilibrium constant, $K _{ c }$ is $3.75 \times 10^{-6}$ at 1069 K . Calculate the $K _{ p }$ for the reaction at this temperature?

Answer

We know that,
$
K_{p}=K_{c}(RT)^{\Delta n}
$
For the above reaction,
$
\begin{aligned}
\Delta n & =(2+1)-2=1 \\
K_{p} & =3.75 \times 10^{-6}(0.0831 \times 1069) \\
K_{p} & =0.033
\end{aligned}
$

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