The p-Block Elements Test

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The p-Block Elements Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statements is wrong?

A
Single N–N bond is stronger than the single P–P bond.

B
\(PH_3\) can act as a ligand in the formation of coordination compound with transition elements.

C
\(NO_2\) is paramagnetic in nature

D
Covalency of nitrogen in \(N_2O_5\) is four.

Solution

N-N bond is weaker than the single P-P bond. because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length.
Question 2
1 / -0

A brown ring is formed in the ring test for \(NO_3^–\) ion. It is due to the formation of

A
\([Fe(H_2O)_5 (NO)]^{2+} \)

B
\(FeSO_4 .NO_2\)

C
\([Fe(H_2O)_4 (NO)_2 ] ^{2+} \)

D
\(FeSO_4 .HNO_3\)

Solution

When freshly prepared solution of \(FeSO_4\) is added in a solution containing \(NO_3^-\) ion, after adding \(H_2SO_4\) from sides of the test tube leads in the development of a brown color complex in between the liquids.

\(NO_3^- + 3Fe^{2+} + 4H^+ \rightarrow\) \(NO + 3Fe^{2+} + 2H_2O\)

\([Fe(H_2O)_6]^{2+} + NO \rightarrow\) \([Fe(H_2O)_5 NO]^{2+} + H_2O\)
Question 3
1 / -0

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
\(Bi_2O_5\)

B
\(BiF_5\)

C
\(BiCl_5\)

D
\(Bi_2 S_5\)

Solution

Stability of +5 state decreases from top to bottom because of the inert pair effect. But because of high electronegativity and smaller size of fluorine bismuth can exist in this form \((BiF_5)\).
Question 4
1 / -0

On heating ammonium dichromate and barium azide separately we get

A
\(N_2\) in both cases

B
\(N_2\) with ammonium dichromate and NO with barium azide

C
\(N_2O\) with ammonium dichromate and \(N_2\) with barium azide

D
\(N_2O\) with ammonium dichromate and \(NO_2\) with barium azide

Solution

On heating ammonium dichromate and barium azide separately we get \( N_2\) gas in both cases.

\((NH_4)_2Cr_2O_7 \overset{\Delta}{\rightarrow} \) \(N_2 + 4H_2O + Cr_2O_3\)

\(Ba(N_3)_2 \rightarrow Ba + 3N_2\)
Question 5
1 / -0

In the preparation of \(HNO_3\), we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of \(NH_3\) will be ______.

A
2

B
3

C
4

D
6

Solution

\(4NH_3 + 5O_2 \underset{Pt}{\stackrel{\Delta}{\rightarrow}} 4NO(g) + 6H_2O\)

Hence, from above equation. oxidation of 2 moles of ammonia will produce 2 moles of NO.
Question 6
1 / -0

The oxidation state of central atom in the anion of compound \(NaH_2 PO_2 \) will be ______.

A
+3

B
+5

C
+1

D
–3

Solution

Oxidation state of \(NaH_2PO_2\)

\(NaH_2PO_2\)

+1 +1 n -2

+1 + 2 × + 1 + x + 2 × -2 = 0

+3 + x -4 = 0

x - 1 = 0

x = +1
Question 7
1 / -0

A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with \(NH_3\) an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________.

A
– 3 to +3

B
– 3 to 0

C
– 3 to +5

D
0 to – 3

Solution

\(MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2\)

\(\overset{\overset{-3}{}}{N}\)\(H_3 + 3Cl_2 \rightarrow \) \(\overset{\overset{+3}{}}{N}\)\(Cl_3 + 3HCl\)

Hence oxidation state of nitrogen changes from -3 to +3.
Question 8
1 / -0

In the preparation of compounds of Xe, Bartlett had taken \(O_2^+ Pt F_6^– \) as a base compound. This is because

A
both \(O_2\) and Xe have same size.

B
both \(O_2\) and Xe have same electron gain enthalpy.

C
both \(O_2\) and Xe have almost same ionisation enthalpy.

D
both Xe and \(O_2\) are gases.

Solution

Neil Bartlett, then observed the reaction of a noble gas. First, he prepared a red compound which is formulated as \(O_2^+ Pt F_6^-\) . He then realised that the first ionisation enthalpy of molecular oxygen (1175 kJ \(mol^{-1}\)) was almost identical with that of xenon (1170 kJ \(mol^{-1}\)).

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The p-Block Elements Test - 15

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The p-Block Elements Test - 15

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SHARING IS CARING

Question 1

Single N–N bond is stronger than the single P–P bond.

\(PH_3\) can act as a ligand in the formation of coordination compound with transition elements.

\(NO_2\) is paramagnetic in nature

Covalency of nitrogen in \(N_2O_5\) is four.

Solution

Question 2

\([Fe(H_2O)_5 (NO)]^{2+} \)

\(FeSO_4 .NO_2\)

\([Fe(H_2O)_4 (NO)_2 ] ^{2+} \)

\(FeSO_4 .HNO_3\)

Solution

Question 3

\(Bi_2O_5\)

\(BiF_5\)

\(BiCl_5\)

\(Bi_2 S_5\)

Solution

Question 4

\(N_2\) in both cases

\(N_2\) with ammonium dichromate and NO with barium azide

\(N_2O\) with ammonium dichromate and \(N_2\) with barium azide

\(N_2O\) with ammonium dichromate and \(NO_2\) with barium azide

Solution

Question 5

2

3

4

6

Solution

Question 6

+3

+5

+1

–3

Solution

Question 7

– 3 to +3

– 3 to 0

– 3 to +5

0 to – 3

Solution

Question 8

both \(O_2\) and Xe have same size.

both \(O_2\) and Xe have same electron gain enthalpy.

both \(O_2\) and Xe have almost same ionisation enthalpy.

both Xe and \(O_2\) are gases.

Solution

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