- ABarium tetrafluorobromate $(V)$
- BBarium tetrafluorobromate $(III)$
- ✓Barium bis (tetrafluorobromate) $(III)$
- Dnone of these
$Z =$ Atomic number of the metal
$X =$ Number of electrons lost during the formation of the metal ion from its atom
$Y = Number$ of electrons donated by the ligands
$(i)$ For $\left.\left[ Ph _3 P \right)_2 PdCl _2\right] AN$ of $Pd =46-2+4 \times 2=52$.
The atomic number of xnon is 54 .
Hence, it does not follows EAN.
$(ii)$ For $[ NiBrCl ( en )], EAN$ of $Ni =28-2+8=34$.
The atomic number of krypton is 36 .
Hence, it does not follows EAN.
$(iii)$ For $Na _4\left[ Fe ( CN )_5 NOS \right], EAN$ of $Fe =26-2+5 \times 2+2=36$.
This is equal to the atomic number of krypton.
Hence, it follows EAN.
$(iv)$ For $Cr ( CO )_3( NO )_2, EAN$ of $Cr =24-0+3 \times 2+3 \times 2=36[ Kr ]$.
This is equal to the atomic number of krypton.
Hence, it follows $EAN.$
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$(a)$ Number of $S-S$ bonds in $H_2S_nO_6$ are $(n +1)$
$(b)$ When $F_2$ reacts with $H_2O$ it forms $HF,$ $O_2$ & $O_3.$
$(c)$ $XeF_6$ on hydrolysis shows disproportionation reaction
$(d)$ $Al$ metal on reacting with dilute $NaOH$ gives a white precipitate of $Al (OH)_3$ as a final product