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प्रश्न
The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.
Based upon the above information, arrange the following orbitals in the increasing order of energy.
5f, 6d, 7s, 7p
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उत्तर
| Orbitals | s | p | d | f |
| l | 0 | 1 | 2 | 3 |
| Orbitals | n | n + l |
| 5f | 5 | 8 |
| 6d | 6 | 8 |
| 7s | 7 | 7 |
| 7p | 7 | 8 |
Thus the increasing order of energy is 7s < 5f < 6d < 7p.
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संबंधित प्रश्न
Choose the correct option.
p-orbitals are _________ in shape.
Choose the correct option.
Principal Quantum number describes -
State Heisenberg uncertainty principle.
State and explain Pauli’s exclusion principle.
Explain the anomalous behaviour of copper.
Explain the anomalous behaviour of chromium.
Write orbital notations for the electron in orbitals with the following quantum numbers.
n = 2, l = 1
Write condensed orbital notation of electronic configuration of the following element:
Oxygen (Z = 8)
Draw shapes of 2s orbitals.
Draw shapes of 2p orbitals.
The electronic configuration of oxygen is written as 1s2 2s2 \[\ce{2p^2_{{x}}}\] \[\ce{2p^1_{{y}}}\] \[\ce{2p^1_{{z}}}\] and not as 1s2 2s2 \[\ce{2p^2_{{x}}}\], \[\ce{2p^2_{{y}}}\] \[\ce{2p^0_{{z}}}\], Explain.
Write a note on ‘Principal Quantum number.
Which one of the following orders is CORRECT in case of energy of the given subshells?
P: n = 4; l = 3
Q: n = 5; I = 1
R: n = 5; l = 0
S: n = 4; l = 2
How many electrons in 19K have n = 3, l = 1?
The three electrons have the following set of quantum numbers:
X = 6, 1, −1, `+1/2`
Y = 6, 0, 0, `+1/2`
Z = 5, 1, 0, `+1/2`
Identify the CORRECT statement.
How many electrons can fit in the orbital for which n = 4 and l = 2?
Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals:
| (i) | (a) `n = 3, l = 2, m_l = -2, m_s = - 1/2` |
| (b) `n = 3, l = 2, m_l = -1, m_s = - 1/2` | |
| (ii) | (a) `n = 3, l = 1, m_l = 1, m_s = + 1/2` |
| (b) `n = 3, l = 2, m_l = 1, m_s = + 1/2` | |
| (iii) | (a) `n = 4, l = 1, m_l = 1, m_s = + 1/2` |
| (b) `n = 3, l = 2, m_l = 1, m_s = + 1/2` | |
| (iv) | (a) `n = 3, l = 2, m_l = +2, m_s = - 1/2` |
| (b) `n = 3, l = 2, m_l = +2, m_s = + 1/2` |
The arrangement of orbitals on the basis of energy is based upon their (n + l) value. Lower the value of (n + l), lower is the energy. For orbitals having same values of (n + l), the orbital with lower value of n will have lower energy.
Based upon the above information, arrange the following orbitals in the increasing order of energy.
4s, 3s, 3p, 4d
