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General Principles and Processes of Isolation of Elements

Extraction of metals

Amalgamation method of extraction of metal is possible because

Question

Easy

Solving time: 2 mins

Amalgamation method of extraction of metal is possible because

of low melting point of mercury

low reactivity of mercury

amalgamated metal is less reactive

mercury is readily available

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Book:

IIT JEE Super Course in Chemistry: Inorganic Chemistry (Pearson)

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Text solutionVerified

When metal comes in contact with mercury, it immediately forms the amalgam.
It is very convenient to get back the metal simply through distillation due to low melting point of mercury.

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Practice questions from IIT JEE Super Course in Chemistry: Inorganic Chemistry (Pearson)

Question 1

Easy

Siderite ore is decomposed in excess of air, the solid product formed is

$F e_{3} O_{4}$
$PbO$
$F e_{2} O_{3}$
$Fe S_{2}$

Topic:

General Principles and Processes of Isolation of Elements

Book:

IIT JEE Super Course in Chemistry: Inorganic Chemistry (Pearson)

View solution

Question 2

Easy

Statement

1

:
In the blast furnace, during extraction of iron, iron melts at a much lower temperature than its melting point.
and
Statement

2

:
Iron produced in the blast furnace is impure in nature.

Statement– $1$ is True, Statement– $2$ is True; Statement– $2$ is a correct explanation for Statement– $1$
Statement– $1$ is True, Statement– $2$ is True; Statement– $2$ is NOT a correct explanation for Statement– $1$
Statement– $1$ is True, Statement– $2$ is False
Statement– $1$ is False, Statement– $2$ is True

Topic:

General Principles and Processes of Isolation of Elements

Book:

IIT JEE Super Course in Chemistry: Inorganic Chemistry (Pearson)

View solution

Question 3

Easy

Why Titanium metal is not prepared by electrolysis of molten salt or carbon reduction of oxide?

Topic:

General Principles and Processes of Isolation of Elements

Book:

IIT JEE Super Course in Chemistry: Inorganic Chemistry (Pearson)

View solution

Question 4

Medium

An ore is subjected to roasting. On doing so, a gas

A

is liberated and the metal is obtained in free state. If gas turns acidified dichromate paper green, the ore is

cassiterite
copper glance
malachite
sphalerite

Topic:

General Principles and Processes of Isolation of Elements

Book:

IIT JEE Super Course in Chemistry: Inorganic Chemistry (Pearson)

View solution

Practice more questions from General Principles and Processes of Isolation of Elements

Question 1

Easy

The metal that is extracted from sea water is:

$Na$
$Ca$
$Mg$
$Sn$

Topic:

General Principles and Processes of Isolation of Elements

View solution

Question 2

Medium

Ag_{2} S + NaCN + Zn ⟶ Ag

This method of extraction of Ag by complex formation and then its displacement is called

Parke's method
Mac Arthur-Forrest method
Serpeck method
Hall's method

Topic:

General Principles and Processes of Isolation of Elements

Book:

New Pattern JEE Problems Chemistry for JEE Main and Advance( Master Problem Package) (Arihant)

View solution

Question 3

Easy

In the extraction of copper, metal is formed in the Bessemer converter due to reaction:

$Cu_{2} S + 2 Cu_{2} O \to 6 Cu + SO_{2}$
$Cu_{2} S \to 2 Cu + S$
$Fe + Cu_{2} O \to 2 Cu + FeO$
$2 Cu_{2} O \to Cu + O_{2}$

Topic:

General Principles and Processes of Isolation of Elements

View 2 solutions

Question 4

Hard

Which one of the following reactions is an example of calcination process ?

$2 Ag + 2 HCl + [O] \to 2 AgCl + H_{2} O$
$2 Zn + O_{2} \to 2 ZnO$
$2 ZnS + 3 O_{2} \to 2 ZnO + 2 SO_{2}$
$MgCO_{3} \to MgO + CO_{2}$

Topic:

General Principles and Processes of Isolation of Elements

View 16 solutions

Practice questions on similar concepts asked by Filo students

Question 1

Thus, the new element first gets a be readily seen that the number of elements in temporary name, with symbol consisting of each period is twice the number of atomic three letters. Later permanent name and orbitals available in the energy level that is symbol are given by a vote of IUPAC being filled. The first period

(n = 1)

starts with representatives from each country. The the filling of the lowest level

(1 s)

and therefore permanent name might reflect the country (or has two elements - hydrogen (

1 s^{'}

) and helium state of the country) in which the element was

(1 s^{2})

when the first shell

(K)

is completed. The discovered, or pay tribute to a notable second period

(n = 2)

staris with lithium and scientist. As of now, elements with atomic the third electron enters the 2 sorbital. The next numbers up to 118 have been discovered. element, beryllium has four electrons and has Official names of all elements have been the electronic configuration

1 s^{2} 2 s^{2}

. Starting announced by IUPAC. from the next element boron, the

2 p

orbitals

Problem 3.1 completed at neon (2 s^{2} 2 p^{6}) . Thus there are are filled with electrons when the L shell is

What would be the IUPAC name and

8

elements in the second period. The third symbol for the element with atomic period

(n = 3)

begins at sodium, and the added number 120 ?

electron enters a

3 s

orbital. Successive filling Solution

3 s

and

3 p

orbitals gives rise to the third From Table 3.4, the roots for 1,2 and 0 period of 8 elements from sodium to argon. The: are un, bi and nil, respectively. Hence, the symbol and the name respectively are Ubn

and unbinilium. filling up of 3 d orbitals becomes energetically you may note that before the 4 p orbital is filled,

favourable and we come across the so called

3 d

transition series of elements. This starts 3.5 ELECTRONIC CONFIGURATIONS OF from scandium

(Z = 21)

which has the electronic ELEMENTS AND THE PERIODIC configuration

3 d^{1} 4 s^{2}

. The

3 d

orbitals are filled TABLE In the preceding unit we have learnt that an at zinc

(Z = 30)

with electronic configuration electron in an atom is characterised by a set of

3 d^{10} 4 s^{2}

. The fourth period ends at krypton four quantum numbers, and the principal with the filling up of the

4 p

orbitals. Altogether quantum number (n) defines the 18 elements in this fourth period. The level known as shell. With period

(n = 5)

beginning with rubidium is We have also studied similar to the fourth period and contains the subshells, in reltals

(s, p, d, (Z = 39)

. This period ends at xenon with the f) in an atom. The distribution of electrons into filling up of the

5 p

orbitals. The sixth period orbitals of an atom is called its electronic

(n = 6)

contains 32 elements and successive configuration. An element's location in the electrons enter

6 s, 4 f, 5 d

and

6 p

orbitals, in Periodic Table reflects the quantum numbers the order - filling up of the

4 f

orbitals begins of the last orbital filled. In this section we will with cerium

(Z = 58)

and ends at lutetium observe a direct connection between the

(Z = 71)

to give the

4 f

-inner transition series electronic configurations of the elements and which is called the lanthanoid series. The the long form of the Periodic Table.

seventh period

(n = 7)

7 s

, The period indiates

5 f, 6 d

and

7 p

orbitals and includes most of The period indicates the value of

n

for the the man-made radioactive elements. This outermost or valence shell. In other words, period will end at the element with atomic successive period in the Periodic Table is number 118 which would belong to the noble associated with the filling of the next higher gas family. Filling up of the

5 f

orbitals after principal energy level

(n = 1, n = 2

, etc.). It can actinium

(Z = 89)

gives the

5 f

-inner transition

Topic:

Periodic Table

View solution

Question 2

A graph between

lo g t_{1/2}

and

lo g a

(abscisas). 'a' is the initial concentration of A in the reaction, is given. For the reaction,

A \to

Product, the rate law is:

A \frac{- d [ A ]}{d t} = K

(B)

\frac{d [ A ]}{d t} = K [A]

Your Arswer - (c)

\frac{d ∣ A ∣}{d t} = K ∣ A]^{2}

D \frac{d [ A ]}{d t} = K [A]^{3}

- Correct Answer-

Topic:

Chemical Kinetics

View solution

Question 3

In Cannizzaro's reaction, the intermediate that will

Topic:

Aldehydes & Ketones

View 2 solutions

Question 4

The following unbalanced reaction takes place in acidic solution:

(12.6, 15.2)

Cr_{2} O_{7}^{2 -} (a q) + Fe^{2 +} (a q) ⟶ Cr^{3 +} (a q) + Fc^{3 +} (a q)

a. Write the balanced equation.b. How many milliliters of a

0.211 M K_{2} Cr_{2} O_{7}

solution are needed to react with

5.00 g

FeSO_{4}

Topic:

Chemistry

View solution

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Question Text	Amalgamation method of extraction of metal is possible because
Topic	General Principles and Processes of Isolation of Elements
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1
Upvotes	116