190024
\(BCl_3\) does not exist as dimer but \(BH_3\) exists as dimer \((B_2H_6)\) because
1 Chlorine is more electronegative than hydrogen
2 there is \(p\pi -p\pi \) back bonding in \(BCl_3\) but \(BH_3\) does not contain such multiple bonding
3 large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms
4 none of the above
Explanation:
Large sized chlorine atoms do not fit in between the small baron atoms whereas small sized hydrogen atoms get fitted in between boron atoms so \(BCl _3\) does not exist as dimer but \(BH _3\) exist as dimer \(\left( B _2 H _6\right)\). \(B _2 H _6+ H _2 O \stackrel{\text { Cold is enough }}{\longrightarrow} H _3 BO _3+6 H _2\)
11. THE P-BLOCK ELEMENTS
190025
\(Ge\,(II)\) compounds are powerful reducing agents whereas \(Pb\,(IV)\) compounds are strong oxidants. It can be due to
1 \(Pb\) is more electropositive than \(Ge\)
2 Ionization potential of lead is less than that of \(Ge\)
3 Ionic radii of \(Pb^{2+}\) and \(Pb^{4+}\) are larger than those of \(Ge^{2+}\) and \(Ge^{4+}\)
4 More pronounced inert pair effect in lead than in \(Ge\)
Explanation:
\(Ge\,(II)\) tends to acquire \(Ge\,(IV)\) state by loss of electrons. Hence it is reducing in nature. \(Pb\,(IV)\) tends to acquire \(Pb\,(II)\) \(O.S\). by gain of electrons. Hence it is oxidising in nature. This is due to inert pair effect.
11. THE P-BLOCK ELEMENTS
190026
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because
1 Carbon is not a powerful reducing agent
2 Magnesium reacts with carbon to form carbides
3 Carbon does not react with magnesium
4 Carbon is a non-metal
Explanation:
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because magnesium reacts with carbon to form carbides. \(2 Mg +3 C \stackrel{2000^{\circ} C }{\longrightarrow} Mg _2 C _3\)
11. THE P-BLOCK ELEMENTS
190027
Carborundum is obtained when silica is heated at high temperature with
1 carbon
2 carbon monoxide
3 carbon dioxide
4 calcium carbonate
Explanation:
Silica on heating with carbon at elevated temperature, gives carborundum (silicon carbide) \(Si{O_2}\, + \,3C\,\xrightarrow{\Delta }\,\mathop {SiC}\limits_{carborundum} \, + \,2CO\) Carborundum is a very hard substance
190024
\(BCl_3\) does not exist as dimer but \(BH_3\) exists as dimer \((B_2H_6)\) because
1 Chlorine is more electronegative than hydrogen
2 there is \(p\pi -p\pi \) back bonding in \(BCl_3\) but \(BH_3\) does not contain such multiple bonding
3 large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms
4 none of the above
Explanation:
Large sized chlorine atoms do not fit in between the small baron atoms whereas small sized hydrogen atoms get fitted in between boron atoms so \(BCl _3\) does not exist as dimer but \(BH _3\) exist as dimer \(\left( B _2 H _6\right)\). \(B _2 H _6+ H _2 O \stackrel{\text { Cold is enough }}{\longrightarrow} H _3 BO _3+6 H _2\)
11. THE P-BLOCK ELEMENTS
190025
\(Ge\,(II)\) compounds are powerful reducing agents whereas \(Pb\,(IV)\) compounds are strong oxidants. It can be due to
1 \(Pb\) is more electropositive than \(Ge\)
2 Ionization potential of lead is less than that of \(Ge\)
3 Ionic radii of \(Pb^{2+}\) and \(Pb^{4+}\) are larger than those of \(Ge^{2+}\) and \(Ge^{4+}\)
4 More pronounced inert pair effect in lead than in \(Ge\)
Explanation:
\(Ge\,(II)\) tends to acquire \(Ge\,(IV)\) state by loss of electrons. Hence it is reducing in nature. \(Pb\,(IV)\) tends to acquire \(Pb\,(II)\) \(O.S\). by gain of electrons. Hence it is oxidising in nature. This is due to inert pair effect.
11. THE P-BLOCK ELEMENTS
190026
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because
1 Carbon is not a powerful reducing agent
2 Magnesium reacts with carbon to form carbides
3 Carbon does not react with magnesium
4 Carbon is a non-metal
Explanation:
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because magnesium reacts with carbon to form carbides. \(2 Mg +3 C \stackrel{2000^{\circ} C }{\longrightarrow} Mg _2 C _3\)
11. THE P-BLOCK ELEMENTS
190027
Carborundum is obtained when silica is heated at high temperature with
1 carbon
2 carbon monoxide
3 carbon dioxide
4 calcium carbonate
Explanation:
Silica on heating with carbon at elevated temperature, gives carborundum (silicon carbide) \(Si{O_2}\, + \,3C\,\xrightarrow{\Delta }\,\mathop {SiC}\limits_{carborundum} \, + \,2CO\) Carborundum is a very hard substance
190024
\(BCl_3\) does not exist as dimer but \(BH_3\) exists as dimer \((B_2H_6)\) because
1 Chlorine is more electronegative than hydrogen
2 there is \(p\pi -p\pi \) back bonding in \(BCl_3\) but \(BH_3\) does not contain such multiple bonding
3 large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms
4 none of the above
Explanation:
Large sized chlorine atoms do not fit in between the small baron atoms whereas small sized hydrogen atoms get fitted in between boron atoms so \(BCl _3\) does not exist as dimer but \(BH _3\) exist as dimer \(\left( B _2 H _6\right)\). \(B _2 H _6+ H _2 O \stackrel{\text { Cold is enough }}{\longrightarrow} H _3 BO _3+6 H _2\)
11. THE P-BLOCK ELEMENTS
190025
\(Ge\,(II)\) compounds are powerful reducing agents whereas \(Pb\,(IV)\) compounds are strong oxidants. It can be due to
1 \(Pb\) is more electropositive than \(Ge\)
2 Ionization potential of lead is less than that of \(Ge\)
3 Ionic radii of \(Pb^{2+}\) and \(Pb^{4+}\) are larger than those of \(Ge^{2+}\) and \(Ge^{4+}\)
4 More pronounced inert pair effect in lead than in \(Ge\)
Explanation:
\(Ge\,(II)\) tends to acquire \(Ge\,(IV)\) state by loss of electrons. Hence it is reducing in nature. \(Pb\,(IV)\) tends to acquire \(Pb\,(II)\) \(O.S\). by gain of electrons. Hence it is oxidising in nature. This is due to inert pair effect.
11. THE P-BLOCK ELEMENTS
190026
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because
1 Carbon is not a powerful reducing agent
2 Magnesium reacts with carbon to form carbides
3 Carbon does not react with magnesium
4 Carbon is a non-metal
Explanation:
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because magnesium reacts with carbon to form carbides. \(2 Mg +3 C \stackrel{2000^{\circ} C }{\longrightarrow} Mg _2 C _3\)
11. THE P-BLOCK ELEMENTS
190027
Carborundum is obtained when silica is heated at high temperature with
1 carbon
2 carbon monoxide
3 carbon dioxide
4 calcium carbonate
Explanation:
Silica on heating with carbon at elevated temperature, gives carborundum (silicon carbide) \(Si{O_2}\, + \,3C\,\xrightarrow{\Delta }\,\mathop {SiC}\limits_{carborundum} \, + \,2CO\) Carborundum is a very hard substance
190024
\(BCl_3\) does not exist as dimer but \(BH_3\) exists as dimer \((B_2H_6)\) because
1 Chlorine is more electronegative than hydrogen
2 there is \(p\pi -p\pi \) back bonding in \(BCl_3\) but \(BH_3\) does not contain such multiple bonding
3 large sized chlorine atoms do not fit in between the small boron atoms where as small sized hydrogen atoms get fitted in between boron atoms
4 none of the above
Explanation:
Large sized chlorine atoms do not fit in between the small baron atoms whereas small sized hydrogen atoms get fitted in between boron atoms so \(BCl _3\) does not exist as dimer but \(BH _3\) exist as dimer \(\left( B _2 H _6\right)\). \(B _2 H _6+ H _2 O \stackrel{\text { Cold is enough }}{\longrightarrow} H _3 BO _3+6 H _2\)
11. THE P-BLOCK ELEMENTS
190025
\(Ge\,(II)\) compounds are powerful reducing agents whereas \(Pb\,(IV)\) compounds are strong oxidants. It can be due to
1 \(Pb\) is more electropositive than \(Ge\)
2 Ionization potential of lead is less than that of \(Ge\)
3 Ionic radii of \(Pb^{2+}\) and \(Pb^{4+}\) are larger than those of \(Ge^{2+}\) and \(Ge^{4+}\)
4 More pronounced inert pair effect in lead than in \(Ge\)
Explanation:
\(Ge\,(II)\) tends to acquire \(Ge\,(IV)\) state by loss of electrons. Hence it is reducing in nature. \(Pb\,(IV)\) tends to acquire \(Pb\,(II)\) \(O.S\). by gain of electrons. Hence it is oxidising in nature. This is due to inert pair effect.
11. THE P-BLOCK ELEMENTS
190026
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because
1 Carbon is not a powerful reducing agent
2 Magnesium reacts with carbon to form carbides
3 Carbon does not react with magnesium
4 Carbon is a non-metal
Explanation:
Carbon cannot be used to produce magnesium by chemical reduction of \(MgO\) because magnesium reacts with carbon to form carbides. \(2 Mg +3 C \stackrel{2000^{\circ} C }{\longrightarrow} Mg _2 C _3\)
11. THE P-BLOCK ELEMENTS
190027
Carborundum is obtained when silica is heated at high temperature with
1 carbon
2 carbon monoxide
3 carbon dioxide
4 calcium carbonate
Explanation:
Silica on heating with carbon at elevated temperature, gives carborundum (silicon carbide) \(Si{O_2}\, + \,3C\,\xrightarrow{\Delta }\,\mathop {SiC}\limits_{carborundum} \, + \,2CO\) Carborundum is a very hard substance
