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# Which of the following is/are true for ${{B}_{2}}$ and ${{C}_{2}}$ molecules according to M.O.T? (This question has multiple correct options).(a)- Both are having $1-\sigma$ and $1-\pi$ bond(b)- Both are having the same bond length.(c)- Both are having different bond orders.(d)-${{B}_{2}}$ is paramagnetic and${{C}_{2}}$ is diamagnetic.

Answer Verified
Hint: If the compound has an unpaired electron then it is paramagnetic, and if it has all paired electrons then it is diamagnetic. We can calculate the bond order of the compound by dividing the sum of electrons in bonding orbital and antibonding orbital with 2.

Complete answer:
Let us study all the options one by one:
(a)- Both are having $1-\sigma$ and $1-\pi$ bond
${{B}_{2}}$ molecule has a single bond between both boron atoms. Hence, it has a $\sigma$ bond.
In ${{C}_{2}}$ molecules there are 2 bonds. One is $\sigma$ bond and the other is $\pi$ bond.
Hence, this option is incorrect.
(b)- Both are having the same bond length.
The bond length of the molecule is related to the bond order of the molecule. Bond length is inversely proportional to the bond order.
In ${{B}_{2}},$ molecule the bond order is 1 and the bond order of ${{C}_{2}}$ is 2. Hence, the bond length of ${{C}_{2}}$ is shorter than ${{B}_{2}}$.
Hence, this option is also incorrect.
(c)- Both are having different bond orders.
The bond order of the compound is calculated by dividing the sum of electrons in bonding orbital and antibonding orbital with 2.
${{B}_{2}}$ has 6 electrons in its outermost shell . Configuration is : $\sigma 2{{s}^{2}}\text{ }\sigma *2{{s}^{2}}\text{ }\pi 2{{p}_{x}}^{1}\text{ }\pi 2{{p}_{y}}^{1}$
Bond order = $\dfrac{b.o-a.b.o}{2}=\dfrac{4-2}{2}=1$
${{C}_{2}}$ has 8 electrons in its outermost shell . Configuration is: $\sigma 2{{s}^{2}}\text{ }\sigma *2{{s}^{2}}\text{ }\pi 2{{p}_{x}}^{2}\text{ }\pi 2{{p}_{y}}^{2}$
Bond order = $\dfrac{b.o-a.b.o}{2}=\dfrac{6-2}{2}=2$
Hence, this option is correct.
(d)- ${{B}_{2}}$ is paramagnetic and ${{C}_{2}}$ is diamagnetic in nature.
${{B}_{2}}$ has two unpaired electrons hence, it is paramagnetic. ${{C}_{2}}$ has all paired electrons hence, it is diamagnetic.
Hence, this is also correct.
So, the correct answer is “Option C and D”.

Note: The orbital $\text{ }\pi 2{{p}_{x}}\text{ and }\pi 2{{p}_{y}}$ gets degenerate after mixing hence they have same energy. So, if there are 2 electrons left each of them gets one-one electron each after that only pairing is done.