Answer
407.4k+ views
Hint Before solving this question, first we have to understand the principles which are mentioned in the question. These principles are used for the filling of electrons in an orbital.
Complete step by step solution:
Aufbau principle states that in the ground state of an atom or ion, the electrons are filled on the basis of the increasing energy levels, this means that the orbital having low energy will be filled first. Pauli’s exclusion principle states that no two electrons have the same set of all the four quantum numbers. And according to the hund's rule of maximum multiplicity, pairing of electrons will be done only when all the degenerate orbitals are singly occupied.
The chemical symbol for iodine is I.
The valence electronic configuration for iodine is $5{{p}^{5}}$ , which means that the last electron will belong to the 5p orbital. Hence, the value of n = 5, l= 1, m= -1,0,1 and the value of spin multiplicity is $\dfrac{1}{2}\text{ or -}\dfrac{1}{2}$
Hence, the correct answer is option (B). i.e. the four quantum numbers (principle, angular, magnetic, spin) for the final electron of iodine is 5, 1, 1, +½
Note: In case when the value of $(n+l)$ is the same for two orbitals, then the orbital having lower value of n i.e. the principal quantum number will be filled first.
Complete step by step solution:
Aufbau principle states that in the ground state of an atom or ion, the electrons are filled on the basis of the increasing energy levels, this means that the orbital having low energy will be filled first. Pauli’s exclusion principle states that no two electrons have the same set of all the four quantum numbers. And according to the hund's rule of maximum multiplicity, pairing of electrons will be done only when all the degenerate orbitals are singly occupied.
The chemical symbol for iodine is I.
The valence electronic configuration for iodine is $5{{p}^{5}}$ , which means that the last electron will belong to the 5p orbital. Hence, the value of n = 5, l= 1, m= -1,0,1 and the value of spin multiplicity is $\dfrac{1}{2}\text{ or -}\dfrac{1}{2}$
Hence, the correct answer is option (B). i.e. the four quantum numbers (principle, angular, magnetic, spin) for the final electron of iodine is 5, 1, 1, +½
Note: In case when the value of $(n+l)$ is the same for two orbitals, then the orbital having lower value of n i.e. the principal quantum number will be filled first.
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