Niobium (Nb) and ruthenium (Ru) have and number of electrons in their respective 4d orbitals. The value of is _______. [2025]

Question

Niobium (Nb) and ruthenium (Ru) have $“ x ”$ and $“ y ”$ number of electrons in their respective 4d orbitals. The value of $x + y$ is _______. [2025]

Smart Achievers · Accepted Answer

(11)

${}_{41}{Nb} = {}_{36}{[Kr]}$ $4 d^{4} 5 s^{1}$

${}_{44}{Ru} = {}_{36}{[Kr]}$ $4 d^{7} 5 s^{1}$

$x + y = 4 + 7 = 11$

Solution

Q.
Niobium (Nb) and ruthenium (Ru) have $“ x ”$ and $“ y ”$ number of electrons in their respective 4d orbitals. The value of $x + y$ is _______. [2025]

Ans.
(11)

${}_{41}{Nb} = {}_{36}{[Kr]}$ $4 d^{4} 5 s^{1}$

${}_{44}{Ru} = {}_{36}{[Kr]}$ $4 d^{7} 5 s^{1}$

$x + y = 4 + 7 = 11$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Niobium (Nb) and ruthenium (Ru) have “x” and “y” number of electrons in their respective 4d orbitals. The value of x+y is _______. [2025]

Ans. (11) Nb41=[Kr]364d45s1 Ru44=[Kr]364d75s1 x+y=4+7=11

Want Us to Email you About Special Offers & Updates?

Q.
Niobium (Nb) and ruthenium (Ru) have $“ x ”$ and $“ y ”$ number of electrons in their respective 4d orbitals. The value of $x + y$ is _______. [2025]

Ans.
(11)

${}_{41}{Nb} = {}_{36}{[Kr]}$ $4 d^{4} 5 s^{1}$

${}_{44}{Ru} = {}_{36}{[Kr]}$ $4 d^{7} 5 s^{1}$

$x + y = 4 + 7 = 11$