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Which of the following statements is not true regarding molecular orbital theory?
Bonding molecular orbital has lower energy than the antibonding molecularorbital.
2s and 2patomic orbitals combine to give how many molecular orbitals?
2s> (σ2s) (0*2s) (2)
2p > (σ2p_{z}) (σ*2p_{z})(π2p_{x})(π*2p_{x}) (π2p_{y}) (π*2p_{y}) (6)
The conditions for the combination of atomic orbitals to form molecular orbitals are stated below. Mark the incorrect condition mentioned here.
Sigma molecular orbitals are symmetrical around the bond axis while pi molecular orbitals are not symmetrical.
The electronic configuration of carbon is Is^{2} 2s^{2} 2p^{2}. There are 12 electrons in C_{2}. The correct electronic configuration of C_{2} molecule is
For atoms like nitrogen and below N in atomic number have higher energy for p_{z} orbital, hence the configuration for C_{2} molecule is
(σls^{2})(σ*1s^{2}) (σ2s^{2}) (σ*2s^{2}) (π2p^{2}_{x }=_{ }π2p^{2}_{y})
The increasing order of energies of various molecular orbitals of N_{2} is given below:
σls <σ*ls <σ2s <σ*2s <π2p_{x} = π2p_{y} <σ2p_{z} < π*2p_{x} = π*2p_{y}<σ*2p_{z}
The above sequence is not true for the molecule
For allel ements which have atomic number more than 7(beyond nitrogen) the energy of σ2p_{z} is lower than
π2p_{x} and π2p_{y} orbitals.
Which of the following species has unpaired electrons?
N_{2}, F_{2} and O^{2}_{2 }have paired electrons
What is the order of stability of N_{2} and its ions?
Bond order of N_{2} = 3, N^{+}_{2} = 2.5 N^{}_{2} = 2.5 and N_{2}^{2} is 2. Higher the bond order, more is the stability.
What will be the bond order of the species with electronic configuration 1s^{2 }2s^{2} 2p^{5}?
Electronic configuration of atom: 1s^{2 }2s^{2} 2p^{5}
M.O. configuration:
Which of the following bond orders is indication of existence of a molecule?
A molecule exists only if the bond order is positive. If bond order is zero or negative, the molecule does not exist.
Which of the following pairs will have same bond order?
F_{2} and O^{2}_{2 }are isoelectronic
Bond order of N^{+}_{2},N^{}_{2} and N_{2} will be
According to molecular orbital theory, which of the following will not exist?
Be_{2}: (σ1s^{2}) (σ*1s^{2}) (σ2s^{2}) (σ*2s^{2})
Since B.O. is zero, the molecule does not exist.
Which of the following facts regarding bond order is not valid?
Higher the bond order, higher is the bond enthalpy.
Which of the following formulae does not show the correct relationship?
Bond order ∝ Bond dissociation energy.
Fill in the blanks with the appropriate choice.
Bond order of N^{+}_{2 } is ___P___ while that of_{ }N_{2} is ___Q___ . Bond order of O^{+}_{2 } is ___R___ while that of O_{2 }is ___S___ . N−N bond distance ___T___ , when N_{2} changes to N^{+}_{2 }and when O_{2 }changes to O^{+}_{2 }, the OO bond distance ___U___ .
Since N^{+}_{2} has lower bond order than N_{2}, bond length of N  N in N^{+}_{2} increases .In O^{+}_{2} , bond order increases form 2 to 2.5 hence , bond length decreases.
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