Question

Chooe the correct amswer using given codes. Use T if statement is true and F if it is false.
(I) Sulphide ions reacts with $N{a}_2\left[Fe\right(CN{)}_5\left(NO\right)]$ to form a purple coloured compound $N{a}_4\left[Fe\right(CN{)}_5\left(NOS\right)]$. In the reaction, the oxidation state of iron changes.
(II) $Pt\left(IV\right)$ compounds are relatively more stable than $Ni\left(IV\right)$ compounds.
(III) The welding of magnesium can be done in the atmosphere of helium.
(IV) $LiAI{H}_4$ on hydrolysis will give $H_2$.

• A. FFTT
• B. FTTT
• C. TFTF
• D. TFTT

Answer 1

Answer: (B)

FTTT

(I) Oxidation state of iron in $N{a}_2\left[Fe\right(CN{)}_5\left(NO\right)]$ is $+3$.

Oxidation state of iron in $N{a}_4\left[Fe\right(CN{)}_5\left(NOS\right)]$ is $+3$.

In this reaction, the oxidation state of iron does not change.

Thus, statement is false ($F$).

(II) Going down in the group, the ionization energy required to remove 4 electrons from $Ni$ is higher than required for formation of $P{t}^{+4}$. Thus $Pt\left(IV\right)$ compounds are more stable and $Ni\left(IV\right)$ compounds do not exist. This is due to the presence of shielding effect of $(n-1)d$ sub shell electrons.

Thus, statement is true ($T$).

(III) The welding of magnesium can be done in the atmosphere of helium. Due to higher ionization potential of $He$ inert gas, it produces hotter arc at higher voltage, providing wide deep bead; this is an advantage for magnesium alloys.

Thus, statement is true ($T$).

(IV) $LiAI{H}_4$ on hydrolysis will give $H_2$. It reacts violently with water by producing hydrogen gas. Hence it should not be exposed to moisture and the reactions are performed in inert and dry atmosphere.

$LiAl{H}_4+4H_{2}O⟶LiOH+Al(OH{)}_3+4H_2$

Thus, statement is true ($T$).