Question

Which is the electron configuration for nobelium (No)?
[Rn]7s25f14
[Rn]7s25f7
[Ne]3s23p7
[Xe]6s25d1

Answer 1

Answer:

[Rn]7s25f14

Explanation:

Answer 2

$[Rn]5f^{14}7s^2$ is the electron configuration for nobelium (No).

Explanation:

• The electronic configuration of an element is the distribution of its electrons in different energy levels around the atomic nucleus.
• It is written with help of an atomic number of an element.

Given:

The element nobelium.

To find:

The electronic configuration of nobelium

Solution:

The atomic number of nobelium is 102.

The number of protons = Atomic number = 102

The number of electrons = Number of protons = 102

There are 102 electrons in atom nobelium and they will be filled according to the Aufbau rule that the electron will be filled in a subshell with lower orbital energy.

The electronic configuration of nobelium :

$[No]=1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^{10}6s^26p^65f^{14}7s^2$

Or we can also write the above electronic configuration as:

$[No]=[Rn]5f^{14}7s^2$

$[Rn]5f^{14}7s^2$ is the electron configuration for nobelium (No).

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