Magnetic moment of `M^(x+)` is `sqrt(24)BM`. The number of unpaired electrons in `M^(x+)` is.

Spin magnetic moment of X^(n+) (Z=26) is sqrt(24) B.M. Hence number of unpaired electrons and value of n respectively are:

Spin magnetic moment of X^(n+) (Z=26) is sqrt(24) B.M. Hence number of unpaired electrons and value of n respectively are:

The magnetic moment of M^(x+) (atomic number = 25 is sqrt(15) BM .The number of unpaired electron and the value of x respectively are

A compound in which a metal ion \(M^{x+}\)(Z=25) has a spin only magnetic moment of \(\sqrt{24}\)BM.The number of unpaired electrons in the compound and the oxidation state of the metal ion are respectively.

Magnetic moment of Mn^(+2) is 5.8 B.M. Hence the number of unpaired electrons presenting the metal ion is

A compound of a metal ion M^(X+)(z=24) has a spin only magnetic moment of sqrt(15)B.M. . The number of unpaired electrons in the compound are

A compound of a metal ion M^(X+)(z=24) has a spin only magnetic moment of sqrt(15)B.M. . The number of unpaired electrons in the compound are

The magnetic moment of M^(x+) (atomic number of M=25) is sqrt(15)BM . The number of unpaired electrons and the value of x respectively are-