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# The titanium (atomic number$= 22$) compound that does not exist is:A: $TiO$B: $Ti{O_2}$C: ${K_2}Ti{O_4}$D: ${K_2}Ti{F_6}$

Last updated date: 10th Sep 2024
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Hint: Charge on an atom is due to valence electrons. An element loses its valence electrons, due to which the net positive charge (due to protons) on an atom increases. Due to which the atom acquires a positive charge. If an atom acquires some electrons in its valence shell then net negative charge (due to extra electrons) increases and the atom acquires negative charge.

When an atom gains or loses electrons it acquires some charge and this charge on an atom is due to valence electrons. This charge on the atom is called oxidation state. If the atom gains electrons the oxidation state is negative and if the atom loses electrons the oxidation state is positive.
In this question we have found the compound among given options which is not possible for titanium. For this first we have to write the electronic configuration of titanium. Atomic number of titanium is $22$. Electronic configuration of titanium is as follows:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^2}$
Titanium can lose four electrons (after losing four electrons it will acquire noble gas configuration that is electronic configuration of titanium will be same as that of argon). As titanium can lose four electrons the possible oxidation states of titanium are $+ 1,{\text{ }} + 2,{\text{ }} + 3{\text{ }}and{\text{ }} + 4$ (titanium is losing its electrons due to which oxidation state will be positive). So, we have to find that compound in which oxidation state of titanium is other than $+ 1,{\text{ }} + 2,{\text{ }} + 3{\text{ }}and{\text{ }} + 4$ as titanium can have only these four oxidation states.
In $TiO$, oxygen is more electronegative than titanium. So, the oxidation state of oxygen will be negative and titanium will be positive. Electronegativity refers to the tendency of an atom to attract a shared pair of electrons. Oxidation state of oxygen is $- 2$ (oxygen needs two electrons to acquire noble gas configuration so it will gain two electrons). As a whole, the compound is neutral. This means the oxidation state of titanium will be $+ 2$ (charge on oxygen is $- 2$ and on compound is zero). $+ 2$ Oxidation state of titanium is possible (explained above) this means $TiO$ exists.
In $Ti{O_2}$, charge on oxygen atoms is $- 4$ (charge on one oxygen atom is $- 2$ but in this compound there are two atoms of oxygen so, the charge will be $- 4$). As a whole compound is neutral, charge on titanium will be $+ 4$ (charge on oxygen is $- 4$ and on compound is zero). $+ 4$ Oxidation state of titanium is possible (explained above) this means $Ti{O_2}$ exists.
In ${K_2}Ti{O_4}$, there are four atoms of oxygen and the charge on these four atoms is $- 8$ (charge on one oxygen atom is $- 2$ but in this compound there are four atoms of oxygen so, the charge will be $- 8$). Charge on one potassium atom is $+ 1$ (potassium needs to lose one electron to acquire noble gas configuration. As it loses electrons it will have positive oxidation state). There are two potassium ions present in this compound. So, the charge due to potassium will be $+ 2$. As a whole, the compound is neutral. So, the oxidation state of titanium will be $+ 6$ (as a whole compound is neutral, charge due to oxygen atoms is $- 8$ and charge due to potassium atoms is $+ 2$). $+ 6$ Oxidation state for titanium is not possible. This means the compound ${K_2}Ti{O_4}$ does not exist.

So, the correct answer is Option C .

Note:
Fluorine is more electronegative than oxygen due to which all the compounds in which oxygen and fluorine are present, the oxidation state of oxygen is $+ 2$ because being more electronegative fluorine will pull electrons from oxygen and oxygen will acquire positive charge as it loses electrons.