JEE Mains · Chemistry · STD 11 - 2. structure of atom
The correct decreasing order of energy, for the orbitals having, following set of quantum numbers. \((A)\) \(n=3,1=0, m=0\) \((B)\) \(n=4,1=0, m=0\) \((C)\) \(n =3,1=1, m =0\) \((D)\) \(n=3,1=2, m=1\)
- A \(( D )>( B )>( C )>( A )\)
- B \((B) >( D )>( C )>( A )\)
- C \(( C )>( B )>( D )>( A )\)
- D \((B) >( C )>( D )>( A )\)
Answer & Solution
Correct Answer
(A) \(( D )>( B )>( C )>( A )\)
Step-by-step Solution
Detailed explanation
\((A)\) \(n+\ell=3+0=3\) \((B)\) \(n+\ell=4+0=4\) \((C)\) \(n+\ell=3+1=4\) \((D)\) \(n+\ell=3+2=5\) Higher \(n+\ell\) value, higher the energy if same \(n +\ell\) value, then higher \(n\) value, higher the energy.Thus: \(D > B > C > A\).
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