In Niobium $(Z = 41)$,the number of electrons with $m = -1$ will be:

  • A
    $7$ or $8$
  • B
    $8$ or $9$
  • C
    $1$ or $2$
  • D
    $3$ or $4$

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Similar Questions

$(i)$ Write the electronic configurations of the following ions:
$(a)$ $H^{-}$
$(b)$ $Na^{+}$
$(c)$ $O^{2-}$
$(d)$ $F^{-}$
$(ii)$ What are the atomic numbers of elements whose outermost electrons are represented by
$(a)$ $3s^{1}$
$(b)$ $2p^{3}$ and
$(c)$ $3p^{5}$?
$(iii)$ Which atoms are indicated by the following configurations?
$(a)$ $[He] 2s^{1}$
$(b)$ $[Ne] 3s^{2} 3p^{3}$
$(c)$ $[Ar] 4s^{2} 3d^{1}$

The number of given orbitals which have electron density along the axis is $..........$. $p_{x}, p_{y}, p_{z}, d_{xy}, d_{yz}, d_{xz}, d_{z^2}, d_{x^2-y^2}$

Assuming Hund's Rule is not necessarily followed for a $d^5$ configuration,what could be the maximum and minimum multiplicity possible?

The maximum number of orbitals which can be identified with $n=4$ and $m_l=0$ is $\qquad$

According to the Aufbau principle,which orbital is filled with electrons immediately after the $5f$ orbital?

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