Cool Chemistry

How will you convert Benzene into the following?

a) p-nitrobenzoic acid

b) m-xylene and

Naming a compound from its molecular formula is definitely an useful skill to master for students studying Chemistry. This guide will help you become familiar with that skill. You can practice these skills and master them.

General Terms

An inorganic compound can either be described as binary, ternary and quarternary. A binary compound is the one that contains two different elements for example KCl; a ternary compound is the one that contains three different elements, for example \(K_2SO_4\) and a quarternary compound is the one that has four elements, for example \(KHSO_3\).

Number Prefixes

The number of atoms in a compound is usually denoted by a prefix. For example in CO, there is an oxygen atom for every carbon atom and hence it is called Carbon Monoxide. “Mono” is the term that indicates one.

Here are the terms that indicate One to Ten. Mono is One; Bi is Two; Tri is Three; Tetra is four; Penta is five; Hexa is Six, Hepta is Seven; Octa is Eight; Nona is Nine and Deca is Ten.

In \(B_2O_3\) there are \(1\frac{1}{2}\) oxygen atoms for every boron atom. This is indicated by the term “sesqui“. So, \(B_2O_3\) is called Boron Sesquioxide.

Examples: \(SO_2\) is Sulphur Dioxide, \(CCl_4\) is Carbon Tetrachloride and \(P_2O_3\) is Phosphorus Sesquioxide.

Metallic Cations

Some metal atoms show a single valency (oxidation state) and some metal atoms show multiple valencies (oxidation states). For example, Magnesium shows +2 oxidation state in all its compounds and Copper can show +1 and +2 oxidation states.

If the metal atom shows only on oxidation state, its name in the cationic state is the name of the metal itself. For example, \(Mg^{2+}\) is called Magnesium ion.

For metals showing multiple oxidation states the naming can be done in any of the following two ways.

(a) The name or the Latin name of the metal ion can me suffixed in such a way that the oxidation state can be identified from it. If the metal is present in its lowest oxidation state its name is suffixed with -ous and  if the metal is present in its highest oxidation state its name is suffixed with -ic. Following are some examples;

Fe is Ferrum. If Fe is present in its lower oxidation state +2, it is called Ferrous ion and if Fe is present in its higher oxidation state, +3, it is called Ferric ion.

Cu is Cuprum. If Cu is present in its lower oxidation state +1, it is called Cuprous ion and if Cu is present in its higher oxidation state, +2, it is called Cupric ion.

Similarly, Mercury is called Mercurous ion in its +1 state and Mercuric ion in its +2 state.

(b) The name of the metal is followed by the oxidation state indicated by Roman Numeral in parentheses.Following are some examples;

\(Cu^+\) can be called Copper (I) ion and \(Cu^{2+}\) can be called Copper (II) ion. Similarly, \(Pb^{2+}\) can be called Lead (II) ion and \(Pb^{4+}\) can be called as Lead (IV) ion.

Anions

The anion in a compound could either be elemental or a group of atoms called radicals. For example, \(Cl^-\) is an elemental anion and \(ClO_3^-\) is a radical.

(a) The names of elemental anions end with the suffix -ide. SO, \(Cl^-\) is called chloride and \(S^{2-}\) is called sulphide.

(b) Typically in an anionic radical the central atom will be connected to few oxygen atoms. If the radical is monovalent (i.e. has an overal charge of -1), it will usually found to be connected to three oxygen atoms. If the radical is polyvalent (i.e. has an overall charge of more than -1), it will usually found to be connected to four oxygen atoms. In both the cases described the name of the anionic radical will end with the suffix -ate. For example, \(ClO_3^-\) is called chlorate and \(PO_4^{3-}\) is called phosphate.

If there are less oxygen atoms than expected, then the name of the anionic radical will end with a different suffix to indicate it. If there is one oxygen atom less than expected, the name of the anionic radical will end with -ite. Examples are \(SO_3^{2-}\) ion - sulphite ion and \(NO_2^-\) ion - nitrite ion.

If there is one oxygen more than expected, the name of the anionic radical gets a prefix -per along with the suffix -ate. For example, \(BrO_4^-\) is called perbromate ion.

If there are two oxygen atoms less than expected, the name of the anionic radical gets a prefix -hypo along with the suffix -ite. For example, \(ClO^-\) is called hypochlorite ion.

(c) If an anion is formed by removal of hydrogen atoms from a molecule (conjugate base of acids), the name of the anionic radical starts with a prefix -bi. For example, \(HS^-\) is called bisulphide ion.

How will you bring about the following conversion?

Volume Strength of \(H_2O_2\) is the volumes of \(O_2\) released on decomposition of one volume of \(H_2O_2\) at STP. so, if a \(H_2O_2\) solution is labelled 10 Vol, decomposing 1L of that solution will give 10L of \(O_2\) and decomposing 1mL of it will give 10mL of \(O_2\).

On heating \(H_2O_2\) decomposes as follows;

Two moles of \(H_2O_2\) on decomposition gives a mole of \(O_2\). Or, in other words, the moles of \(O_2\) formed is half the moles of \(H_2O_2\) decomposed.

If we take 1 mole \(H_2O_2\) solution and decompose it completely it will release 0.5 mole of \(O_2\). At STP one mole of any gas occupies 22.4L. So, 0.5 mole of \(O_2\) will occupy \(22.4 L/mole \times 0.5 mole = 11.2L\).

If the molarity of a \(H_2O_2\) solution is M, 1L of this solution will have M moles of \(H_2O_2\) in it and 1L of it on decomposition will give \(\frac{M}{2}\) moles of \(O_2\). \(\frac{M}{2}\) moles of \(O_2\) will occupy \(22.4 L/mole \times \frac{M}{2} mole = 11.2M L\). This 11.2M L of \(O_2\) has come from 1L of \(H_2O_2\) solution of molarity M.

So, if the molarity of a \(H_2O_2\) solution is M, its volume strength will be 11.2M.

Oleum also known as fuming sulphuric acid has the molecular formula (\(H_2S_2O_7\)). It can be considered as a mixture od \(SO_3\) and \(H_2SO_4\). Addition of water to Oleum converts the free \(SO_3\) into \(H_2SO_4\) and the resulting solution will contain only \(H_2SO_4\).

\[SO_3 + H_2O \rightarrow H_2SO_4\]

To know how much water to add to a known amount of oleum, the concentration of oleum is expressed in terms of a percentage. The percentage will look like (100 + x)% and will always be more than 100. The value of “x“ is the amount of \(H_2O\) required by 100g of oleum to convert all the free \(SO_3\) present in it into \(H_2SO_4\).

For example, if a sample of oleum is labelled 109%, it means that to 100g of it 9g of water should be added to convert all the free \(SO_3\) present in it into \(H_2SO_4\).

\[SO_3 + H_2O \rightarrow H_2SO_4\]

The product of the reaction given below is

The reaction of propene with HOCl (\(Cl_2 + H_2O\)) proceeds through the intermediate

A) \(CH_3-CH^+-CH_2-Cl\)

B) \(CH_3-CH(OH)-CH_2^+\)

C) \(CH_3-CHCl-CH_2^+\)

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure \(p_i\) and temperature \(T_1\) are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to \(T_2\). The final pressure \(p_f\) is

A) \(2p_i (\dfrac{T_1}{T_1 + T_2})\)

B) \(2p_i (\dfrac{T_2}{T_1 + T_2})\)

C) \(2p_i (\dfrac{T_1T_2}{T_1 + T_2})\)

In the Hofmann bromamide degradation reaction, the number of moles of NaOH and \(Br_2\) used per mole of amine produced are

A) Four moles of NaOH and two moles of \(Br_2\)

B) Two moles of NaOH and two moles of \(Br_2\)

C) Four moles of NaOH and one mole of \(Br_2\)

Which of the following atoms has the highest first ionization energy?

A) Na

B) K

C) Sc

Galvanization is applying a coating of

A) Cr

B) Cu

C) Zn

Thiol Group is present in

A) Cystine

B) Cysteine

C) Methionine