3 \(\mathrm{O}_2^{+}\)have higher bond order than \(\mathrm{O}_2^{-}\).
4 \(\mathrm{O}_2^{+}\)is diamagnetic while \(\mathrm{O}_2^{-}\)is paramagnetic.
Explanation:
(d) Molecular orbital configuration of \(\mathrm{O}_2^{+}\):
\(\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma * 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2\)
\(=\pi 2 p_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^1\)
Bond order \(=\frac{1}{2}(10-5)=2.5\)
Number of unpaired electrons \(=1\), hence paramagnetic.
Molecular orbital configuration of \(\mathrm{O}_2^{-}\):
\(\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2\)
\(=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^1\)
Bond order \(=\frac{1}{2}(10-7)=1.5\)
Number of unpaired electron \(=1\), hence paramagnetic.
Both \(\mathrm{O}_2^{+}\)and \(\mathrm{O}_2^{-}\)are paramagnetic.
\(\mathrm{O}_2^{+}\)have higher bond order than \(\mathrm{O}_2^{-}\).
\(\mathrm{O}_2^{-}\)is less stable.
Karnataka CET 2015
CHEMISTRY(KCET)
285268
The pair of compounds having identical shapes for their molecules is
1 \(\mathrm{CH}_4, \mathrm{SF}_4\)
2 \(\mathrm{BCl}_2, \mathrm{ClF}_3\)
3 \(\mathrm{XeF}_2, \mathrm{ZnCl}_2\)
4 \(\mathrm{SO}_2, \mathrm{CO}_2\)
Explanation:
(c)\(\mathrm{XeF}_2\) is linear in shape with 2 bond pairs and 3 lone pairs of electrons.
\(\mathrm{ZnCl}_2\) is also linear in shape with 2 bond pairs of electrons.
3 \(\mathrm{O}_2^{+}\)have higher bond order than \(\mathrm{O}_2^{-}\).
4 \(\mathrm{O}_2^{+}\)is diamagnetic while \(\mathrm{O}_2^{-}\)is paramagnetic.
Explanation:
(d) Molecular orbital configuration of \(\mathrm{O}_2^{+}\):
\(\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma * 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2\)
\(=\pi 2 p_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^1\)
Bond order \(=\frac{1}{2}(10-5)=2.5\)
Number of unpaired electrons \(=1\), hence paramagnetic.
Molecular orbital configuration of \(\mathrm{O}_2^{-}\):
\(\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2\)
\(=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^1\)
Bond order \(=\frac{1}{2}(10-7)=1.5\)
Number of unpaired electron \(=1\), hence paramagnetic.
Both \(\mathrm{O}_2^{+}\)and \(\mathrm{O}_2^{-}\)are paramagnetic.
\(\mathrm{O}_2^{+}\)have higher bond order than \(\mathrm{O}_2^{-}\).
\(\mathrm{O}_2^{-}\)is less stable.
Karnataka CET 2015
CHEMISTRY(KCET)
285268
The pair of compounds having identical shapes for their molecules is
1 \(\mathrm{CH}_4, \mathrm{SF}_4\)
2 \(\mathrm{BCl}_2, \mathrm{ClF}_3\)
3 \(\mathrm{XeF}_2, \mathrm{ZnCl}_2\)
4 \(\mathrm{SO}_2, \mathrm{CO}_2\)
Explanation:
(c)\(\mathrm{XeF}_2\) is linear in shape with 2 bond pairs and 3 lone pairs of electrons.
\(\mathrm{ZnCl}_2\) is also linear in shape with 2 bond pairs of electrons.
3 \(\mathrm{O}_2^{+}\)have higher bond order than \(\mathrm{O}_2^{-}\).
4 \(\mathrm{O}_2^{+}\)is diamagnetic while \(\mathrm{O}_2^{-}\)is paramagnetic.
Explanation:
(d) Molecular orbital configuration of \(\mathrm{O}_2^{+}\):
\(\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma * 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2\)
\(=\pi 2 p_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^1\)
Bond order \(=\frac{1}{2}(10-5)=2.5\)
Number of unpaired electrons \(=1\), hence paramagnetic.
Molecular orbital configuration of \(\mathrm{O}_2^{-}\):
\(\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2\)
\(=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^1\)
Bond order \(=\frac{1}{2}(10-7)=1.5\)
Number of unpaired electron \(=1\), hence paramagnetic.
Both \(\mathrm{O}_2^{+}\)and \(\mathrm{O}_2^{-}\)are paramagnetic.
\(\mathrm{O}_2^{+}\)have higher bond order than \(\mathrm{O}_2^{-}\).
\(\mathrm{O}_2^{-}\)is less stable.
Karnataka CET 2015
CHEMISTRY(KCET)
285268
The pair of compounds having identical shapes for their molecules is
1 \(\mathrm{CH}_4, \mathrm{SF}_4\)
2 \(\mathrm{BCl}_2, \mathrm{ClF}_3\)
3 \(\mathrm{XeF}_2, \mathrm{ZnCl}_2\)
4 \(\mathrm{SO}_2, \mathrm{CO}_2\)
Explanation:
(c)\(\mathrm{XeF}_2\) is linear in shape with 2 bond pairs and 3 lone pairs of electrons.
\(\mathrm{ZnCl}_2\) is also linear in shape with 2 bond pairs of electrons.