P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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311.
Elements of group - 15 form compounds in + 5 oxidation state. However, bismuth forms only one well characterised compound in + 5 oxidation
state. The compound is
A
$$B{i_2}{O_5}$$
B
$$Bi{F_5}$$
C
$$BiC{l_5}$$
D
$$B{i_2}{S_5}$$
Answer :
$$Bi{F_5}$$
The stability of + 5 oxidation state decreases down the group due to inert pair effect. The only well characterised $$Bi\left( V \right)$$ compound is $$Bi{F_5}$$ as fluorine being most electronegative
element is able to unpair $$ns$$ electrons.
312.
If chlorine is passed through a solution of hydrogen sulphide in water, the solution turns turbid due to the formation of
A
free chlorine
B
free sulphur
C
nascent oxygen
D
nascent hydrogen
Answer :
free sulphur
\[{{H}_{2}}S+C{{l}_{2}}\xrightarrow{\text{water}}2HCl+S\]
Due to formation of free sulphur the solution becomes turbid.
313.
What happens when diborane reacts with Lewis bases?
A
It forms boron trihydride $$\left( {B{H_3}} \right)$$ due to cleavage.
B
It undergoes cleavage to give borane adduct $$B{H_3} \cdot L$$ ( where, $$L =$$ Lewis base ).
C
It oxidises to give $${B_2}{O_3}.$$
D
It does not react with Lewis bases.
Answer :
It undergoes cleavage to give borane adduct $$B{H_3} \cdot L$$ ( where, $$L =$$ Lewis base ).
A
The stability ofhydride increases from $$N{H_3}\,{\text{to}}\,Bi{H_3}$$ in group $$15$$ of the periodic table.
B
Nitrogen cannot form $$d\pi - p\pi $$ bond.
C
Single$$N - N$$ bond is weaker than the single $$P - P$$ bond.
D
$${N_2}{O_4}$$ has two resonance structures.
Answer :
The stability ofhydride increases from $$N{H_3}\,{\text{to}}\,Bi{H_3}$$ in group $$15$$ of the periodic table.
The ease of formation and stability of hydrides decreases rapidly from $$N{H_3}\,{\text{to}}\,Bi{H_3}.$$ This is evident from their dissociation temperature which decreases from $$N{H_3}\,{\text{to}}\,Bi{H_3}.$$ As we go down the group the size of central atom increases and thus metal-hydrogen bond becomes weaker due to decreased overlap between the large central atom and small hydrogen atom.
$$\mathop {N{H_3} > P{H_3}}\limits_{{\text{(most stable)}}} > As{H_3} > Sb{H_3} > \mathop {Bi{H_3}}\limits_{{\text{(least stable}})} $$
315.
Maximum bond angle at nitrogen is present in which of the following?
A
$$N{O_2}$$
B
$$NO_2^ - $$
C
$$NO_2^ + $$
D
$$NO_3^ - $$
Answer :
$$NO_2^ + $$
Species
Hybridisation
Bond angle
$$N{O_2}$$
$$sp$$
less than $${120^ \circ }$$
$$NO_2^ - $$
$$s{p^2}$$
$${115.4^ \circ }$$
$$NO_2^ + $$
$$sp$$ (linear)
$${180^ \circ }$$
$$NO_3^ - $$
$$s{p^2}$$
$${120^ \circ }$$
So, $$NO_2^ + $$ has maximum bond angle.
316.
Aqueous solution of $$N{a_2}{S_2}{O_3}$$ on reaction with $$C{l_2}$$ gives —
A
$$N{a_2}{S_4}{O_6}$$
B
$$NaHS{O_4}$$
C
$$NaCl$$
D
$$NaOH$$
Answer :
$$NaHS{O_4}$$
The following reaction occurs
$$N{a_2}{S_2}{O_3} + 4C{l_2} + 5{H_2}O\,\,\, \to \,\,2NaHS{O_4} + 8HCl.\,\,$$
317.
$$Pb{F_4},PbC{l_4}$$ exist but $$PbB{r_4}$$ and $$Pb{I_4}$$ do not exist because of
A
large size of $$B{r^ - }$$ and $${I^ - }$$
B
strong oxidising character of $$P{b^{4 + }}$$
C
strong reducing character of $$P{b^{4 + }}$$
D
low electronegativity of $$B{r^ - }$$ and $${I^ - }.$$
Answer :
strong oxidising character of $$P{b^{4 + }}$$
$$F$$ and $$Cl$$ are more oxidising in nature and can achieve $$Pb$$ in $$\left( {IV} \right)O.S.$$ but $$B{r_2}$$ and $${I_2}$$ can not achieve $$Pb$$ in $$\left( {IV} \right)O.S.$$ secondly $$P{b^{4 + }}$$ is strong in oxidising nature and in its presence, $$B{r^ - }$$ and $${I^ - }$$ can not exist.
318.
Least thermally stable is –
A
$$CC{l_4}$$
B
$$SiC{l_4}$$
C
$$GeC{l_4}$$
D
$$GeB{r_4}$$
Answer :
$$GeB{r_4}$$
The thermal stability of tetrahalides decreases in order $$C{X_4} > Si{X_4} > Ge{X_4} > Sn{X_4}$$ and in terms of same metal with different halides is in order of $$M{F_4} > MC{l_4} > MB{r_4} > M{I_4}.$$
320.
A group 14 element is oxidised to form corresponding oxide which is gaseous in nature, when dissolved in water $$pH$$ of the water decreases further addition of group 2 hydroxides leads to precipitation. This oxide can be
A
$$Ge{O_2}$$
B
$$CO$$
C
$$C{O_2}$$
D
$$Sn{O_2}$$
Answer :
$$C{O_2}$$
$$C{O_2}$$ forms carbonic acid $${H_2}C{O_3},$$ when dissolved in water, $$CO$$ is neutral, whereas other two $$Ge{O_2}$$ and $$Sn{O_2}$$ are solids.