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QNEET-2023-Ch-212: NEET 2023 Chemistry - Class 11 NEET-2023-Ch-212 Solution: The correct order of energies of molecular orbitals of \(N_2\)…

The correct order of energies of molecular orbitals of N2N_2 molecule is σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2py)<σ∗2pz\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<σ∗2pz<(π2px=π2py)<(π∗2px=π∗2py)\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \sigma^* 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) σ1s<σ∗1s<σ2s<σ∗2s<(π2px=π2py)<(π∗2px=π∗2py)<σ2pz<σ∗2pz\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma 2p_z < \sigma^* 2p_z σ1s<σ∗1s<σ2s<σ∗2s<(π2px=π2py)<σ2pz<(π∗2px=π∗2py)<σ∗2pz\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z

Multiple Choice Options

  • A \sigma 1 \mathrm{~s} &lt; \sigma^{*} 1 \mathrm{~s} &lt; \sigma 2 \mathrm{~s} &lt; \sigma^{*} 2 \mathrm{~s} &lt; \sigma 2 \mathrm{p}_{\mathrm{z}} &lt; \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) &lt; \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right) &lt; \sigma^{*} 2 \mathrm{p}_{\mathrm{z}} _ 1   s < _ _ 1   s < _ 2   s < _ _ 2   s < _ 2 p z _ < ( � 2 p x _ = � 2 p y _ ) < ( � _ 2 p x _ = � _ 2 p y _ ) < _ _ 2 p z _
  • B \sigma 1 \mathrm{~s} &lt;\sigma^{*} 1 \mathrm{~s} &lt;\sigma 2 \mathrm{~s} &lt; \sigma^{*} 2 \mathrm{~s} &lt; \sigma 2 \mathrm{p}_{\mathrm{z}} &lt; \sigma^{*} 2 \mathrm{p}_{\mathrm{z}} &lt; \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) &lt; \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right) _ 1   s < _ _ 1   s < _ 2   s < _ _ 2   s < _ 2 p z _ < _ _ 2 p z _ < ( � 2 p x _ = � 2 p y _ ) < ( � _ 2 p x _ = � _ 2 p y _ )
  • C \sigma 1 \mathrm{~s} &lt;\sigma^{*} 1 \mathrm{~s} &lt;\sigma 2 \mathrm{~s} &lt; \sigma^{*} 2 \mathrm{~s} &lt; \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) &lt; \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right) &lt; \sigma 2 \mathrm{p}_{\mathrm{z}} &lt; \sigma^{*} 2 \mathrm{p}_{\mathrm{z}} _ 1   s < _ _ 1   s < _ 2   s < _ _ 2   s < ( � 2 p x _ = � 2 p y _ ) < ( � _ 2 p x _ = � _ 2 p y _ ) < _ 2 p z _ < _ _ 2 p z _
  • D \sigma 1 \mathrm{~s} &lt;\sigma^{*} 1 \mathrm{~s} &lt;\sigma 2 \mathrm{~s} &lt; \sigma^{*} 2 \mathrm{~s} &lt; \left(\pi 2 \mathrm{p}_{\mathrm{x}}=\pi 2 \mathrm{p}_{\mathrm{y}}\right) &lt; \sigma 2 \mathrm{p}_{\mathrm{z}} &lt; \left(\pi^{*} 2 \mathrm{p}_{\mathrm{x}}=\pi^{*} 2 \mathrm{p}_{\mathrm{y}}\right) &lt; \sigma^{*} 2 \mathrm{p}_{\mathrm{z}} _ 1   s < _ _ 1   s < _ 2   s < _ _ 2   s < ( � 2 p x _ = � 2 p y _ ) < _ 2 p z _ < ( � _ 2 p x _ = � _ 2 p y _ ) < _ _ 2 p z _

Step-by-Step Verified Solution

Molecular orbital (energy) diagram / sequence of N 2

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