Magnetic moment of complex [Pd(NH3)2Cl2] will be

The correct answer is 0.

Let's determine the magnetic moment of the complex $\left[\mathrm{Pd}{\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]$ step-by-step:

Step 1: Find the oxidation state of palladium (Pd) in the complex.
Let the oxidation state of Pd be $x$.
Ammonia (${\mathrm{NH}}_3$) is a neutral ligand (charge = 0).
Chloride (${\mathrm{Cl}}^{-}$) is an anionic ligand (charge = -1).
Since the overall complex is neutral:
$x+2\left(0\right)+2(-1)=0$
$x-2=0$
$x=+2$
Thus, the oxidation state of palladium is +2, represented as ${\mathrm{Pd}}^{2+}$.

Step 2: Determine the electronic configuration of ${\mathrm{Pd}}^{2+}$.
Palladium is a second-row transition metal (4d series) located in Group 10, with atomic number Z = 46.
The ground-state electronic configuration of neutral Pd is $\left[\mathrm{Kr}\right]4d^{10}$.
Therefore, the configuration of the ${\mathrm{Pd}}^{2+}$ ion is:
${\mathrm{Pd}}^{2+}\Rightarrow \left[\mathrm{Kr}\right]4d^8$

Step 3: Analyze the effect of the ligands.
For 4d and 5d transition metal series elements, the crystal field splitting energy (${\Delta }_o$ or ${\Delta }_{\mathrm{sp}}$) is exceptionally large due to the greater spatial extension of 4d/5d orbitals. Consequently, all ligands (including weak-field ligands like ${\mathrm{Cl}}^{-}$ as well as ${\mathrm{NH}}_3$) behave as strong-field ligands (SFL) and induce pairing of electrons.

Step 4: Determine the hybridization and number of unpaired electrons.
Due to the strong-field behavior of the ligands in this 4d complex, the eight electrons in the 4d subshell pair up completely in the four lower-energy d-orbitals:
$4d^8\Rightarrow (\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )()$
This leaves one 4d orbital empty, allowing for ${\mathrm{dsp}}^2$ hybridization, which gives the complex a square planar geometry.
The number of unpaired electrons ($n$) is:
$n=0$

Step 5: Calculate the magnetic moment.
The spin-only magnetic moment ($\mu$) is calculated using the formula:
$\mu =\sqrt{n(n+2)}B.M.$
Substituting $n=0$ into the equation:
$\mu =\sqrt{0(0+2)}=0B.M.$

Since there are no unpaired electrons, the complex $\left[\mathrm{Pd}{\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]$ is diamagnetic, and its magnetic moment is 0.