
What is the electronic configuration for Boron?
(a) \[1{{s}^{2}}2{{s}^{2}}\]
(b) \[2{{s}^{2}}2{{p}^{1}}\]
(c) \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\]
(d) \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\]
Answer
232.8k+ views
Hint: Electronic configuration of elements is the way in which the electrons fill up in different orbitals of an atom. Fill up the orbitals by keeping in mind the following rules – Aufbau principle, Pauli’s exclusion principle and Hund’s rule.
Complete step by step answer:
Let us try to understand the process of filling electrons by the following principles –
1) Aufbau principle:
It states that, “in the ground state of an atom, an electron enters the orbital of lowest energy first and subsequent electrons are fed in the order of increasing energies”.
Therefore, orbitals fill in the following order –
1s, 2s, 2p, 3p, 4s, 3d, 4p, 5s, 6s, 4f, 5d, 6p, 7s
2) Pauli’s Exclusion principle
It states that, “any orbital can accommodate a maximum of two electrons”.
3) Hund’s rule of maximum multiplicity
It states that, “electron pairing will not take place in orbitals of same energy (same subshell) until each orbital is singly filled.”
Following these rules, let us try to write the electronic configuration of Boron.
Atomic number of Boron is 5, therefore we can say it has five electrons. The wide electronic configuration of Boron will be – 2,3. The first shell contains 2 electrons and the second shell contains 3 electrons.
Now filling up these shells we get – \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\].
Therefore, the answer is – option (d).
Additional Information:
Bohr Bury’s rule is used to decide the energy of the orbitals, i.e. orbitals fill in increasing order of (n+l).
Note: You can also write the answer directly if you know the atomic number of the element.Boron has an atomic number equal to 5.
Therefore, looking at the options we can see that –
A. \[1{{s}^{2}}2{{s}^{2}}\]- 4 electrons are present – incorrect option.
B. \[2{{s}^{2}}2{{p}^{1}}\]- 3 electrons are present – incorrect option.
C. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\]- 10 electrons are present – incorrect option.
D. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\]- 5 electrons are present – correct option.
Complete step by step answer:
Let us try to understand the process of filling electrons by the following principles –
1) Aufbau principle:
It states that, “in the ground state of an atom, an electron enters the orbital of lowest energy first and subsequent electrons are fed in the order of increasing energies”.
Therefore, orbitals fill in the following order –
1s, 2s, 2p, 3p, 4s, 3d, 4p, 5s, 6s, 4f, 5d, 6p, 7s
2) Pauli’s Exclusion principle
It states that, “any orbital can accommodate a maximum of two electrons”.
3) Hund’s rule of maximum multiplicity
It states that, “electron pairing will not take place in orbitals of same energy (same subshell) until each orbital is singly filled.”
Following these rules, let us try to write the electronic configuration of Boron.
Atomic number of Boron is 5, therefore we can say it has five electrons. The wide electronic configuration of Boron will be – 2,3. The first shell contains 2 electrons and the second shell contains 3 electrons.
Now filling up these shells we get – \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\].
Therefore, the answer is – option (d).
Additional Information:
Bohr Bury’s rule is used to decide the energy of the orbitals, i.e. orbitals fill in increasing order of (n+l).
Note: You can also write the answer directly if you know the atomic number of the element.Boron has an atomic number equal to 5.
Therefore, looking at the options we can see that –
A. \[1{{s}^{2}}2{{s}^{2}}\]- 4 electrons are present – incorrect option.
B. \[2{{s}^{2}}2{{p}^{1}}\]- 3 electrons are present – incorrect option.
C. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\]- 10 electrons are present – incorrect option.
D. \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\]- 5 electrons are present – correct option.
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