316657
On doping Ge metal with a little of In, one gets
1 p-type semiconductor
2 n-type semiconductor
3 Insulator
4 Rectifier
Explanation:
Doping of gp. 13 element (In) with Ge (gp. 14 element) cause p-type semiconductor. Doping of gp. 15 element (As) with Ge (gp. 14 element) causes n-type semiconductor.
CHXI11:THE P-BLOCK ELEMENTS
316658
Match Column I with Column II and choose the correct combination from the options given.
\(\mathrm{BF}_{4}^{-} \rightarrow \mathrm{sp}^{3}\) - tetrahedral \(\mathrm{AlCl}_{3} \rightarrow\) Lewis acid due to electron deficiency. \(\mathrm{SnO} \rightarrow\) Can be oxidised to \(\mathrm{SnO}_{2}\) \(\mathrm{PbO}_{2} \Rightarrow \mathrm{Pb}\) in +4 oxidation state and changes to +2 oxidation state. So, acts as oxidising agent.
CHXI11:THE P-BLOCK ELEMENTS
316659
Find the incorrect match.
1 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{6}}}\;\;\;{\mkern 1mu} {\kern 1pt} \;\;\;\;{\mkern 1mu} - \;\;\;{\mkern 1mu} {\kern 1pt} 3{\rm{c}} - 4{\rm{e}}\) bond is present
2 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_6}\;{\mkern 1mu} - \;\) All carbon atoms are \({\rm{s}}{{\rm{p}}^{\rm{3}}} - \) hybridized
Boron in \(\mathrm{B}(\mathrm{OH})_{4}^{-}\)and \(\mathrm{B}_{2} \mathrm{H}_{6}\) is \(\mathrm{sp}^{3}\) hybridised. Silicon in \(\mathrm{SiO}_{4}^{-4}\) is \(\mathrm{sp}^{3}\) hybridised. Carbon in fullurenes is \(\mathrm{sp}^{2}\) hybridised. Al in \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybrised.
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CHXI11:THE P-BLOCK ELEMENTS
316657
On doping Ge metal with a little of In, one gets
1 p-type semiconductor
2 n-type semiconductor
3 Insulator
4 Rectifier
Explanation:
Doping of gp. 13 element (In) with Ge (gp. 14 element) cause p-type semiconductor. Doping of gp. 15 element (As) with Ge (gp. 14 element) causes n-type semiconductor.
CHXI11:THE P-BLOCK ELEMENTS
316658
Match Column I with Column II and choose the correct combination from the options given.
\(\mathrm{BF}_{4}^{-} \rightarrow \mathrm{sp}^{3}\) - tetrahedral \(\mathrm{AlCl}_{3} \rightarrow\) Lewis acid due to electron deficiency. \(\mathrm{SnO} \rightarrow\) Can be oxidised to \(\mathrm{SnO}_{2}\) \(\mathrm{PbO}_{2} \Rightarrow \mathrm{Pb}\) in +4 oxidation state and changes to +2 oxidation state. So, acts as oxidising agent.
CHXI11:THE P-BLOCK ELEMENTS
316659
Find the incorrect match.
1 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{6}}}\;\;\;{\mkern 1mu} {\kern 1pt} \;\;\;\;{\mkern 1mu} - \;\;\;{\mkern 1mu} {\kern 1pt} 3{\rm{c}} - 4{\rm{e}}\) bond is present
2 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_6}\;{\mkern 1mu} - \;\) All carbon atoms are \({\rm{s}}{{\rm{p}}^{\rm{3}}} - \) hybridized
Boron in \(\mathrm{B}(\mathrm{OH})_{4}^{-}\)and \(\mathrm{B}_{2} \mathrm{H}_{6}\) is \(\mathrm{sp}^{3}\) hybridised. Silicon in \(\mathrm{SiO}_{4}^{-4}\) is \(\mathrm{sp}^{3}\) hybridised. Carbon in fullurenes is \(\mathrm{sp}^{2}\) hybridised. Al in \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybrised.
316657
On doping Ge metal with a little of In, one gets
1 p-type semiconductor
2 n-type semiconductor
3 Insulator
4 Rectifier
Explanation:
Doping of gp. 13 element (In) with Ge (gp. 14 element) cause p-type semiconductor. Doping of gp. 15 element (As) with Ge (gp. 14 element) causes n-type semiconductor.
CHXI11:THE P-BLOCK ELEMENTS
316658
Match Column I with Column II and choose the correct combination from the options given.
\(\mathrm{BF}_{4}^{-} \rightarrow \mathrm{sp}^{3}\) - tetrahedral \(\mathrm{AlCl}_{3} \rightarrow\) Lewis acid due to electron deficiency. \(\mathrm{SnO} \rightarrow\) Can be oxidised to \(\mathrm{SnO}_{2}\) \(\mathrm{PbO}_{2} \Rightarrow \mathrm{Pb}\) in +4 oxidation state and changes to +2 oxidation state. So, acts as oxidising agent.
CHXI11:THE P-BLOCK ELEMENTS
316659
Find the incorrect match.
1 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{6}}}\;\;\;{\mkern 1mu} {\kern 1pt} \;\;\;\;{\mkern 1mu} - \;\;\;{\mkern 1mu} {\kern 1pt} 3{\rm{c}} - 4{\rm{e}}\) bond is present
2 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_6}\;{\mkern 1mu} - \;\) All carbon atoms are \({\rm{s}}{{\rm{p}}^{\rm{3}}} - \) hybridized
Boron in \(\mathrm{B}(\mathrm{OH})_{4}^{-}\)and \(\mathrm{B}_{2} \mathrm{H}_{6}\) is \(\mathrm{sp}^{3}\) hybridised. Silicon in \(\mathrm{SiO}_{4}^{-4}\) is \(\mathrm{sp}^{3}\) hybridised. Carbon in fullurenes is \(\mathrm{sp}^{2}\) hybridised. Al in \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybrised.
316657
On doping Ge metal with a little of In, one gets
1 p-type semiconductor
2 n-type semiconductor
3 Insulator
4 Rectifier
Explanation:
Doping of gp. 13 element (In) with Ge (gp. 14 element) cause p-type semiconductor. Doping of gp. 15 element (As) with Ge (gp. 14 element) causes n-type semiconductor.
CHXI11:THE P-BLOCK ELEMENTS
316658
Match Column I with Column II and choose the correct combination from the options given.
\(\mathrm{BF}_{4}^{-} \rightarrow \mathrm{sp}^{3}\) - tetrahedral \(\mathrm{AlCl}_{3} \rightarrow\) Lewis acid due to electron deficiency. \(\mathrm{SnO} \rightarrow\) Can be oxidised to \(\mathrm{SnO}_{2}\) \(\mathrm{PbO}_{2} \Rightarrow \mathrm{Pb}\) in +4 oxidation state and changes to +2 oxidation state. So, acts as oxidising agent.
CHXI11:THE P-BLOCK ELEMENTS
316659
Find the incorrect match.
1 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{6}}}\;\;\;{\mkern 1mu} {\kern 1pt} \;\;\;\;{\mkern 1mu} - \;\;\;{\mkern 1mu} {\kern 1pt} 3{\rm{c}} - 4{\rm{e}}\) bond is present
2 \({\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}} \right)_6}\;{\mkern 1mu} - \;\) All carbon atoms are \({\rm{s}}{{\rm{p}}^{\rm{3}}} - \) hybridized
Boron in \(\mathrm{B}(\mathrm{OH})_{4}^{-}\)and \(\mathrm{B}_{2} \mathrm{H}_{6}\) is \(\mathrm{sp}^{3}\) hybridised. Silicon in \(\mathrm{SiO}_{4}^{-4}\) is \(\mathrm{sp}^{3}\) hybridised. Carbon in fullurenes is \(\mathrm{sp}^{2}\) hybridised. Al in \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}\) is \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybrised.
