Question

1 g of titanium (at.wt. 48) reacts with $\text{ C}{{\text{l}}_{\text{2}}}\text{ }$ to give $\text{ 3}\text{.21875 g }$ compound. The valency of titanium is ____________

Answer 1

Hint: Valency is known as the combining capacity of an atom or molecule. This is the number of valence electrons which need to be added or removed such that the atom attains the noble gas configuration. Titanium reacts with chlorine the general reaction is as shown below, $\text{ Ti + }\dfrac{y}{2}\text{ C}{{\text{l}}_{\text{2}}}\text{ }\to \text{ TiC}{{\text{l}}_{\text{y}}}\text{ }$
Where y is the valency of titanium.

Complete step by step solution:
We are given the following data:
1 gram Weight of titanium which reacts with chlorine.
The weight of the product given is $\text{ 3}\text{.21875 g }$ .
The atomic weight of titanium is 48
To find the valency of titanium.
Let's consider a reaction of titanium and chlorine. Titanium and chlorine react with each other to form a titanium chloride. The general reaction of titanium with y moles of chlorine gives the one mole of $\text{ TiC}{{\text{l}}_{\text{y}}}\text{ }$, where y can be treated as the valency of the titanium. the system is expressed as,
$\text{ Ti + }\dfrac{y}{2}\text{ C}{{\text{l}}_{\text{2}}}\text{ }\to \text{ TiC}{{\text{l}}_{\text{y}}}\text{ }$ (1)
We know that chlorine atom requires one electron to attain the nearest noble gas configuration. Thus the valency of chlorine is $\text{ }-1\text{ }$ .
The general molecular weight of the $\text{ TiC}{{\text{l}}_{\text{y}}}\text{ }$would-be equal to the weight of the titanium and chlorine times the valency of the titanium. that is, $\text{ TiC}{{\text{l}}_{\text{y}}}\text{ }=\text{ }\left( 48\text{ + 35}\text{.5y} \right)\text{ }$
From reaction (1), we know that 48 g of titanium gives out $\text{ }\left( 48\text{ + 35}\text{.5y} \right)\text{ }$ a gram of the product.
Thus we can say that,
48 g of titanium give $\text{ }\left( 48\text{ + 35}\text{.5y} \right)\text{ }$of product,
$\text{ 48 g Ti = }\left( 48\text{ + 35}\text{.5y} \right)\text{ g product }$
Then, 1 gram of titanium would give,
$\text{ 1 g of Ti = x g of product }$
On cross multiplying, the 1 gram of titanium would give, $\text{ }\dfrac{\text{ }\left( 48\text{ + 35}\text{.5y} \right)\text{ }}{48}$ g of product.
Since we know that the total product obtained is$\text{ 3}\text{.21875 g }$. Thus the value of the ‘y’ can be obtained as follows,
$\begin{align}
& \text{ }\dfrac{\text{ }\left( 48\text{ + 35}\text{.5y} \right)\text{ }}{48}\text{ = 3}\text{.21875 } \\
& \Rightarrow \left( 48\text{ + 35}\text{.5y} \right)\text{ = 3}\text{.21875 }\times \text{ }48 \\
& \Rightarrow \text{35}\text{.5y = 154}\text{.5 }-48 \\
& \Rightarrow \text{y = }\dfrac{106.5}{35.5} \\
& \therefore \text{y = + 3} \\
\end{align}$
Thus the value of y is 3. The reaction (1) can be written as,
$\text{ Ti + }\dfrac{3}{2}\text{ C}{{\text{l}}_{\text{2}}}\text{ }\to \text{ TiC}{{\text{l}}_{3}}\text{ }$

Therefore, the valency of titanium in the product is equal $\text{ +3 }$.

Note: Note that, titanium trichloride $\text{ TiC}{{\text{l}}_{\text{3}}}\text{ }$is a very useful catalyst.it is known as the Ziegler-Natta catalyst .it is a stereospecific catalyst and used in the polymerization of propylene to make the polymer polypropylene. Titanium also exhibits the $\text{ +4 }$oxidation state.