
The electronic configuration of transition elements is:
(i) $n{s^1}$
(ii) $n{s^2}n{p^5}$
(iii) $n{s^2}\left( {n - 1} \right){d^{10}}$
(iv) $\left( {n - 1} \right){d^{1 - 10}}n{s^{0 - 2}}$
Answer
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Hint: Electronic configuration is the arrangement of electrons in energy levels around an atomic nucleus. The energy levels are filled following Aufbau’s Principle which states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels.
Complete step-by-step answer:The electron configuration is the distribution of electrons of an atom in energy levels around the nucleus of the atom.
An electron shell described the set of allowed states or energies that share the same principal quantum number $n$ , that electrons may occupy. An atom's ${n^{th}}$ electron shell can accommodate $2{n^2}$ electrons, e.g. the first shell can accommodate two electrons, the second shell eight electrons, the third shell can accommodate up to eighteen electrons and so on.
The shells further consist of subshells which are defined by the azimuthal quantum number$l$. The value of $l$ ranges from $0$ to $\left( {n - 1} \right)$. The values $l\, = \,0,\,1\,,\,2,\,3$ correspond to the $s,\,p,\,d$and $f$labels, respectively having energies $s < p < d < f$. In a subshell the maximum number of electrons that can be placed is $2\left( {2l + 1} \right)$. So two electrons are present in $s$ subshell, six electrons are present in $p$ subshell, ten electrons are present in $d$ subshell and fourteen electrons are present in $f$ subshell.
As stated by Aufbau’s Principle electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, hence following the order: $1s,\,2s,\,2p,\,3s,\,3p,\,4s,\,3d,\,4p,\,5s,......$as orbitals with a lower $\left( {n + l} \right)$value are filled before those with higher $\left( {n + l} \right)$ values. In the case of equal $\left( {n + l} \right)$values, the orbital with a lower $n$ value is filled first.
The elements whose atoms have a partially filled $d$ subshell, or which can give rise to cations with an incomplete $d$ subshell, and belong to the $d - $block elements are known as transition elements. Hence have a general electronic configuration of: $\left( {n - 1} \right){d^{1 - 10}}n{s^{0 - 2}}$.
Hence the correct answer is (iv).
Note:For this kind of question you must have a good knowledge of the placement of different elements in the periodic table. If you know the periodic table you will know the atomic number of the particular element and hence can easily determine the electronic configuration of the element.
Complete step-by-step answer:The electron configuration is the distribution of electrons of an atom in energy levels around the nucleus of the atom.
An electron shell described the set of allowed states or energies that share the same principal quantum number $n$ , that electrons may occupy. An atom's ${n^{th}}$ electron shell can accommodate $2{n^2}$ electrons, e.g. the first shell can accommodate two electrons, the second shell eight electrons, the third shell can accommodate up to eighteen electrons and so on.
The shells further consist of subshells which are defined by the azimuthal quantum number$l$. The value of $l$ ranges from $0$ to $\left( {n - 1} \right)$. The values $l\, = \,0,\,1\,,\,2,\,3$ correspond to the $s,\,p,\,d$and $f$labels, respectively having energies $s < p < d < f$. In a subshell the maximum number of electrons that can be placed is $2\left( {2l + 1} \right)$. So two electrons are present in $s$ subshell, six electrons are present in $p$ subshell, ten electrons are present in $d$ subshell and fourteen electrons are present in $f$ subshell.
As stated by Aufbau’s Principle electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels, hence following the order: $1s,\,2s,\,2p,\,3s,\,3p,\,4s,\,3d,\,4p,\,5s,......$as orbitals with a lower $\left( {n + l} \right)$value are filled before those with higher $\left( {n + l} \right)$ values. In the case of equal $\left( {n + l} \right)$values, the orbital with a lower $n$ value is filled first.
The elements whose atoms have a partially filled $d$ subshell, or which can give rise to cations with an incomplete $d$ subshell, and belong to the $d - $block elements are known as transition elements. Hence have a general electronic configuration of: $\left( {n - 1} \right){d^{1 - 10}}n{s^{0 - 2}}$.
Hence the correct answer is (iv).
Note:For this kind of question you must have a good knowledge of the placement of different elements in the periodic table. If you know the periodic table you will know the atomic number of the particular element and hence can easily determine the electronic configuration of the element.
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