" Magnetic moment of "M^(2+)" is "sqrt(24)BM" .The number of unpaired electrons in is."

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:

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.

Magentic moment of Fe^(a+)(Z= 26 ) is sqrt 24 BM . Hence muber of unpaired electron and value lf a respectively 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

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-