In a diatomic molecule, when combining atomic orbitals, both bonding and anti-bonding molecular orbitals are formed. For atomic orbitals 2s and 2p:
Total number of anti-bonding orbitals:
Total anti-bonding orbitals = 1 (from 2s) + 1 (from σ2p*) + 2 (from 2 π2p*) = 4.
This total, 4, falls within the given range (4,4).
In molecular orbital theory, the anti-bonding molecular orbitals formed from atomic orbitals are as follows:
- Anti-bonding molecular orbital from 2s: 1
- Anti-bonding molecular orbitals from 2p: 3
Thus, the total number of anti-bonding molecular orbitals is:
\(1 + 3 = 4\)
The Correct Answer is: 4
Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
