P - Block Elements MCQ Questions & Answers in Inorganic Chemistry | Chemistry

\[\begin{align} & \left( \text{A} \right){{\left( N{{H}_{4}} \right)}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }C{{r}_{2}}{{O}_{7}}+{{N}_{2}}+{{H}_{2}}O \\ & \left( \text{B} \right)N{{H}_{4}}Cl+NaN{{O}_{2}}\xrightarrow{\Delta }N{{H}_{4}}N{{O}_{2}}+NaCl \\ & \left( \text{C} \right)2N{{H}_{4}}Cl+CaO\xrightarrow{\Delta }CaC{{l}_{2}}+2N{{H}_{3}}\uparrow +{{H}_{2}}O \\ & \left( \text{D} \right)N{{H}_{4}}N{{O}_{2}}\xrightarrow{\Delta }{{N}_{2}}\uparrow +2{{H}_{2}}O \\ & \,\,\,\,\,\,\,\,\,\,Ba{{\left( {{N}_{3}} \right)}_{2}}\xrightarrow{\Delta }Ba+3{{N}_{2}}\downarrow \\ \end{align}\]

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$$Al + NaOH + {H_2}O \to NaAl{O_2} + {H_2} \uparrow .$$

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\[SiC{{l}_{4}}+{{H}_{2}}O\to \underset{\left( X \right)}{\mathop{Si{{\left( OH \right)}_{4}}}}\,\]      \[\xrightarrow{\text{Heat}}\underset{\left( Y \right)}{\mathop{Si{{O}_{2}}}}\,\xrightarrow{NaOH}\underset{\left( Z \right)}{\mathop{N{{a}_{2}}Si{{O}_{3}}}}\,\]

Water gas is produced when steam is passed over red hot coke beds.
$$C\left( s \right) + {H_2}O\left( g \right) \to \underbrace {CO\left( g \right) + {H_2}\left( g \right)}_{{\text{Water gas}}}$$

Oxidation state of $$H$$  is + 1 and that of $$O$$  is - 2.
Let the oxidation state of $$P$$  in the given compounds is $$x.$$
$$\eqalign{ & {\text{In}}\,\,{H_4}{P_2}{O_5} \cr & \left( { + 1} \right) \times 4 + 2 \times x + \left( { - 2} \right) \times 5 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 2x - 10 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 6 \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 3 \cr & {\text{In}}\,\,{H_4}{P_2}{O_6} \cr & \left( { + 1} \right) \times 4 + 2 \times x + \left( { - 2} \right) \times 6 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 2x - 12 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 8 \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 4 \cr & {\text{In}}\,\,{H_4}{P_2}{O_7} \cr & \left( { + 1} \right) \times 4 + 2 \times x + \left( { - 2} \right) \times 7 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 2x - 14 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 10 \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 5 \cr} $$
Thus, the oxidation states of $$P$$  in $${H_4}{P_2}{O_5},{H_4}{P_2}{O_6}$$    and $${H_4}{P_2}{O_7}$$  are + 3, + 4 and + 5 respectively.

The chain length of silicone polymer can be controlled by adding $${\left( {C{H_3}} \right)_3}SiCl,$$   which blocks the ends.

$$CuS{O_4} + 4N{H_3} \to \left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$$        Blue complex due to $$Cu\left( {N{H_3}} \right)_4^{2 + }$$