Question

(a) The effective atomic number of iron in ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$ is?
(b) An excess of $AgN{{O}_{3}}$ is added to $100ml$ of $0.01M$ solution of tetra aqua dichloro chromium $\left( III \right)$ Chloride, the number of moles of $AgCl$ precipitated would be?

Answer 1

(a) Effective atomic number (EAN) in a complex is characterized as the total number of electrons present around the central metal particle.
(b) To solve this part, we must know what molarity indicates- the number of moles of solute per liter of solution (moles/Liter) and is one of the most well-known units used to quantify the concentration of a solution. Molarity can be utilized to compute the volume of solvent or the amount of solute.

Complete step by step solution:
Atomic number $\left( Z \right)-$Oxidation number $+2\times$Coordination number
Complex compound here is ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$
In ferrocyanide ion, ${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$
$Fe\xrightarrow{-2{{e}^{-}}}F{{e}^{2+}}+6C{{N}^{-}}\to {{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$
Formula for EAN $=Z-X+Y$
The atomic number of the central metal atom \left( Fe \right)$$$$Z=26
Number of electrons $\left( X \right)$lost due to the oxidation of $Fe$to F{{e}^{2+}}$$$$=2
Number of electrons$\left( Y \right)$donated by $6C{{N}^{-}}=6\times 2=12.$
Hence, EAN $=26-2+12=36$
Therefore, the correct answer is $36$
(b) The formula for dichloro chromium$\left( III \right)$ chloride is $\left[ Cr\left( {{H}_{2}}O \right)C{{l}_{2}} \right]Cl$
$\left[ Cr\left( {{H}_{2}}O \right)C{{l}_{2}} \right]Cl\xrightarrow{excessAgN{{O}_{3}}}{{\left[ Cr{{\left( {{H}_{2}}O \right)}_{4}}{{\left( Cl \right)}_{2}} \right]}^{+}}+AgCl\downarrow$
Initial $\;\;\;\;\;$ $100\times 0.01$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ $0$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ 0
After ionization 0 $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ 1mol $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ 1mol
One mole of $C{{l}^{-}}$ions react with only $1mol$ of $AgN{{O}_{3}}$molecules to produce $1mol$of$AgCl$
Using Formula, Molarity $=\dfrac{No.\;of\;moles}{V}\times 1000$
$0.01=\dfrac{No\;.of\;moles}{100}\times 1000$
Number of moles of $AgCl=0.001$

Hence, the correct answer is $0.001$.

Note: (a) Since known as the EAN rule, that in various metal complexes the metal atom will in general to surround itself with adequate ligands that the subsequent effective atomic number is mathematically equivalent to the atomic number of the noble gas component found in a similar period in which the metal is located. And this rule is followed by${{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}$.
(b) A mole fraction demonstrates the number of chemical elements. One mole of any substance is equivalent to the value of Avogadro number. It tends to be utilized to measure the products obtained from the chemical reaction. The unit is indicated by mol.