313766
Assertion : \({{\text{B}}_{\text{2}}}\) molecule is diamagnetic Reason : The highest occupied molecular orbital is of \(\sigma\) type.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
The total number of electrons in \({{\text{B}}_{\text{2}}}{\text{ = 10}}\) Molecular orbital configuration: \(\sigma 1 {{\text{s}}^2},\mathop \sigma \limits^* 1 {{\text{s}}^2},\sigma 2 {{\text{s}}^2},\mathop \sigma \limits^* 2 {{\text{s}}^2},\pi 2{\text{p}}_{\text{x}}^1 \approx \pi 2{\text{p}}_{\text{y}}^1\) It has two unpaired electrons. \(\therefore\) It is paramagnetic and the highest occupied molecular orbital is of \(\pi\) type. Hence, both \(\mathrm{A}\) and \(\mathrm{R}\) are incorrect
AIIMS - 2005
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313768
In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?
For NO Total no. of electrons = 15 B.O = 2.5 Magnetic Behaviour = Paramagnetic For \({\rm{N}}{{\rm{O}}^{\rm{ + }}}\) Total no. of electrons = 14 B.O = 3 Magnetic Behaviour = Diamagnetic
JEE - 2013
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313769
According to molecular orbital theory, which of the following will not be a viable molecule?
1 \({\rm{He}}_{\rm{2}}^{\rm{ + }}\)
2 \({\rm{H}}_{\rm{2}}^{\rm{ - }}\)
3 \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\)
4 \({\rm{He}}_{\rm{2}}^{\,\,{\rm{2 + }}}\)
Explanation:
The electronic configuration of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) \( \Rightarrow \left( {{\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}} \right){\rm{,}}{\left( {{{\rm{\sigma }}^{\rm{*}}}{\rm{1s}}} \right)^{\rm{2}}}\) Bond order of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}{\rm{ = }}\frac{{{{\rm{N}}_{\rm{b}}}{\rm{ - }}{{\rm{N}}_{\rm{a}}}}}{{\rm{2}}}{\rm{ = }}\frac{{{\rm{2 - 2}}}}{{\rm{2}}}{\rm{ = 0}}\) Hence \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) does not exist
313766
Assertion : \({{\text{B}}_{\text{2}}}\) molecule is diamagnetic Reason : The highest occupied molecular orbital is of \(\sigma\) type.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
The total number of electrons in \({{\text{B}}_{\text{2}}}{\text{ = 10}}\) Molecular orbital configuration: \(\sigma 1 {{\text{s}}^2},\mathop \sigma \limits^* 1 {{\text{s}}^2},\sigma 2 {{\text{s}}^2},\mathop \sigma \limits^* 2 {{\text{s}}^2},\pi 2{\text{p}}_{\text{x}}^1 \approx \pi 2{\text{p}}_{\text{y}}^1\) It has two unpaired electrons. \(\therefore\) It is paramagnetic and the highest occupied molecular orbital is of \(\pi\) type. Hence, both \(\mathrm{A}\) and \(\mathrm{R}\) are incorrect
AIIMS - 2005
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313768
In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?
For NO Total no. of electrons = 15 B.O = 2.5 Magnetic Behaviour = Paramagnetic For \({\rm{N}}{{\rm{O}}^{\rm{ + }}}\) Total no. of electrons = 14 B.O = 3 Magnetic Behaviour = Diamagnetic
JEE - 2013
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313769
According to molecular orbital theory, which of the following will not be a viable molecule?
1 \({\rm{He}}_{\rm{2}}^{\rm{ + }}\)
2 \({\rm{H}}_{\rm{2}}^{\rm{ - }}\)
3 \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\)
4 \({\rm{He}}_{\rm{2}}^{\,\,{\rm{2 + }}}\)
Explanation:
The electronic configuration of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) \( \Rightarrow \left( {{\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}} \right){\rm{,}}{\left( {{{\rm{\sigma }}^{\rm{*}}}{\rm{1s}}} \right)^{\rm{2}}}\) Bond order of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}{\rm{ = }}\frac{{{{\rm{N}}_{\rm{b}}}{\rm{ - }}{{\rm{N}}_{\rm{a}}}}}{{\rm{2}}}{\rm{ = }}\frac{{{\rm{2 - 2}}}}{{\rm{2}}}{\rm{ = 0}}\) Hence \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) does not exist
313766
Assertion : \({{\text{B}}_{\text{2}}}\) molecule is diamagnetic Reason : The highest occupied molecular orbital is of \(\sigma\) type.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
The total number of electrons in \({{\text{B}}_{\text{2}}}{\text{ = 10}}\) Molecular orbital configuration: \(\sigma 1 {{\text{s}}^2},\mathop \sigma \limits^* 1 {{\text{s}}^2},\sigma 2 {{\text{s}}^2},\mathop \sigma \limits^* 2 {{\text{s}}^2},\pi 2{\text{p}}_{\text{x}}^1 \approx \pi 2{\text{p}}_{\text{y}}^1\) It has two unpaired electrons. \(\therefore\) It is paramagnetic and the highest occupied molecular orbital is of \(\pi\) type. Hence, both \(\mathrm{A}\) and \(\mathrm{R}\) are incorrect
AIIMS - 2005
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313768
In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?
For NO Total no. of electrons = 15 B.O = 2.5 Magnetic Behaviour = Paramagnetic For \({\rm{N}}{{\rm{O}}^{\rm{ + }}}\) Total no. of electrons = 14 B.O = 3 Magnetic Behaviour = Diamagnetic
JEE - 2013
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313769
According to molecular orbital theory, which of the following will not be a viable molecule?
1 \({\rm{He}}_{\rm{2}}^{\rm{ + }}\)
2 \({\rm{H}}_{\rm{2}}^{\rm{ - }}\)
3 \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\)
4 \({\rm{He}}_{\rm{2}}^{\,\,{\rm{2 + }}}\)
Explanation:
The electronic configuration of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) \( \Rightarrow \left( {{\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}} \right){\rm{,}}{\left( {{{\rm{\sigma }}^{\rm{*}}}{\rm{1s}}} \right)^{\rm{2}}}\) Bond order of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}{\rm{ = }}\frac{{{{\rm{N}}_{\rm{b}}}{\rm{ - }}{{\rm{N}}_{\rm{a}}}}}{{\rm{2}}}{\rm{ = }}\frac{{{\rm{2 - 2}}}}{{\rm{2}}}{\rm{ = 0}}\) Hence \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) does not exist
313766
Assertion : \({{\text{B}}_{\text{2}}}\) molecule is diamagnetic Reason : The highest occupied molecular orbital is of \(\sigma\) type.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
The total number of electrons in \({{\text{B}}_{\text{2}}}{\text{ = 10}}\) Molecular orbital configuration: \(\sigma 1 {{\text{s}}^2},\mathop \sigma \limits^* 1 {{\text{s}}^2},\sigma 2 {{\text{s}}^2},\mathop \sigma \limits^* 2 {{\text{s}}^2},\pi 2{\text{p}}_{\text{x}}^1 \approx \pi 2{\text{p}}_{\text{y}}^1\) It has two unpaired electrons. \(\therefore\) It is paramagnetic and the highest occupied molecular orbital is of \(\pi\) type. Hence, both \(\mathrm{A}\) and \(\mathrm{R}\) are incorrect
AIIMS - 2005
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313768
In which of the following ionisation process the bond energy has increased and also the magentic behaviour has changed from paramagnetic to diamagnetic?
For NO Total no. of electrons = 15 B.O = 2.5 Magnetic Behaviour = Paramagnetic For \({\rm{N}}{{\rm{O}}^{\rm{ + }}}\) Total no. of electrons = 14 B.O = 3 Magnetic Behaviour = Diamagnetic
JEE - 2013
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313769
According to molecular orbital theory, which of the following will not be a viable molecule?
1 \({\rm{He}}_{\rm{2}}^{\rm{ + }}\)
2 \({\rm{H}}_{\rm{2}}^{\rm{ - }}\)
3 \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\)
4 \({\rm{He}}_{\rm{2}}^{\,\,{\rm{2 + }}}\)
Explanation:
The electronic configuration of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) \( \Rightarrow \left( {{\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}} \right){\rm{,}}{\left( {{{\rm{\sigma }}^{\rm{*}}}{\rm{1s}}} \right)^{\rm{2}}}\) Bond order of \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}{\rm{ = }}\frac{{{{\rm{N}}_{\rm{b}}}{\rm{ - }}{{\rm{N}}_{\rm{a}}}}}{{\rm{2}}}{\rm{ = }}\frac{{{\rm{2 - 2}}}}{{\rm{2}}}{\rm{ = 0}}\) Hence \({\rm{H}}_{\rm{2}}^{{\rm{2 - }}}\) does not exist
