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25 Questions MCQ Test Chemistry Class 11 - Test: Classification of Elements & Periodicity in Properties - 1

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Test: Classification of Elements & Periodicity in Properties - 1 - Question 1

f block of the Periodic Table consists of

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 1

Lanthanides, Ce(Z = 58) ñ Lu(Z = 71) and Actinoids comprises of f block elements.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 2

The elements charecterised by the filling of 4 f-orbitals, are:

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 2

Elements characterized by the filling of 4f orbitals are known as lanthanides or lanthanoids. These elements belong to the f-block of the periodic table and include:

  • Cerium (Ce)
  • Praseodymium (Pr)
  • Neodymium (Nd)
  • Promethium (Pm)
  • Samarium (Sm)
  • Europium (Eu)
  • Gadolinium (Gd)
  • Terbium (Tb)
  • Dysprosium (Dy)
  • Holmium (Ho)
  • Erbium (Er)
  • Thulium (Tm)
  • Ytterbium (Yb)
  • Lutetium (Lu)

The term 'lanthanides' specifically refers to this series, ensuring clarity in understanding their position in the periodic table.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 3

14 elements after actinium is called

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 3

The correct answer is Option B.

14 elements after actinium is called Actinides.
 

Test: Classification of Elements & Periodicity in Properties - 1 - Question 4

An element belongs to Group 15 and third period of the periodic table. Its electronic configuration will be

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 4

group is 15 so total valence electrons will be 5 while period is 3rd so n value of valence shell will be 3.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 5

It is now recognized that the ‘Modern Periodic Law’ is essentially the consequence of the

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 5

Modern periodic law is essentially the result of periodic variation and electronic configuration.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 6

Metalloids show the properties of

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 6

Metalloids exhibit properties that are intermediate between metals and nonmetals. They share some characteristics with metals, such as:

  • Being solid at room temperature
  • Having a metallic appearance

However, they also exhibit nonmetallic properties, including:

  • Poor electrical conductivity compared to metals

Examples of metalloids include:

  • Boron
  • Silicon
  • Germanium

These elements display this dual nature in their physical and chemical behaviours.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 7

One of the following options is not used for explaining atomic radius

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 7

The term coordinate radius is not typically used to explain atomic radius. Instead, the following terms are relevant:

  • Covalent Radius: Half the distance between the nuclei of two identical atoms when they are covalently bonded.
  • Metallic Radius: The radius of an atom in a metallic lattice, measured from the nucleus to the midpoint of the atoms.
  • Van der Waals’ Radius: Refers to the radius of an atom as determined by the van der Waals force, relevant for noble gases and non-bonding interactions.

Thus, coordinate radius is not recognised as a standard term for explaining atomic radius.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 8

Among the alkali metals cesium is the most reactive because

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 8

Among the alkali metals cesium is the most reactive because the outermost electron is more loosely bound than the outermost electron of the other alkali metals.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 9

In which one of the given formulae of xenon compounds there are five α−bonds and three π-bonds in it?

 

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 9

Test: Classification of Elements & Periodicity in Properties - 1 - Question 10

In the modern periodic table

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 10

In the modern periodic table, the horizontal rows are known as periods, and the vertical columns are referred to as groups.

This structure organises elements based on their:

  • Atomic number
  • Chemical properties

In this context:

  • Periods represent the principal quantum number.
  • Groups indicate similar valence electron configurations.
Test: Classification of Elements & Periodicity in Properties - 1 - Question 11

Which of the following statements is false?

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 11

Alkali metals typically form ionic bonds with non-metals like oxygen, resulting in ionic compounds such as oxides (e.g., Na2O). They do not generally form covalent bonds with oxygen.

Therefore, statement A is false because alkali metals primarily form ionic bonds when reacting with oxygen, not covalent ones.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 12

The order of Decreasing radius is

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 12

The correct order of decreasing radius is:

  • Van der Waals' radius > Metallic radius > Covalent radius.

Explanation:

  • Van der Waals' Radius: This is the largest as it includes the distance between two atoms when they are in contact, encompassing both covalent or metallic radii and additional space due to electron clouds.
  • Metallic Radius: Smaller than Van der Waals', it measures the radius of an atom in a metallic crystal structure, where electrons are delocalised but not as tightly packed as in covalent bonds.
  • Covalent Radius: The smallest, as it measures the distance from the nucleus to the bonding electron pair shared with another atom in a covalent bond.

Thus, the correct order is Van der Waals' radius > Metallic radius > Covalent radius.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 13

For alkali metals, which one of the following trends is INCORRECT?

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 13

Density: Li < Na < K < Rb.
The density of K is lower than that of Na. Thus, option D is incorrect. The correct trend is
Li < K < Na < Rb

Test: Classification of Elements & Periodicity in Properties - 1 - Question 14

Dmitri Mendeleev (1834-1907) and the German chemist, Lothar Meyer (1830-1895) proposed arranging elements in

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 14

Dmitri Mendeleev and Lothar Meyer proposed arranging elements in 1. increasing order of their atomic weights. This arrangement eventually led to the development of the periodic table of elements.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 15

According to the recommendation of International Union of Pure and Applied Chemistry (IUPAC), the groups in the modern periodic table are numbered from

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 15

The groups in the modern periodic table are numbered from 1 to 18 according to IUPAC recommendations. This includes:

  • Group 1: Alkali metals
  • Group 18: Noble gases
Test: Classification of Elements & Periodicity in Properties - 1 - Question 16

Ionization enthalpy increases across a period because

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 16

as we move from left to right in a period atomic radii decreases and due to effective nuclear charge outermost electrons are held more closely to the nucleus.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 17

Choose one of the following in th increasing order of bond length

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 17
  1. N2​: Triple bond → Shortest bond length.
  2. O2: Double bond → Shorter bond length than single bonds.
  3. F2: Single bond, but shorter than Clbecause fluorine atoms are smaller.
  4. Cl2​: Single bond → Longest bond length due to larger atomic size.

Correct Order (Increasing Bond Length): 

Test: Classification of Elements & Periodicity in Properties - 1 - Question 18

For the four successive transition elements (Cr, Mn, Fe and Co), the stability of + 2 oxidation state will be there in which of the following order? (At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 18

Mn has most stable +2 stable because Mn+2 had d5 configuration which is most stable.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 19

A transition metal ion exists in its highest oxidation state. It is expected to behave as

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 19

When a transition metal ion is in highest oxidation state it tends to lower its oxidation state by gaining electrons. thus it itself reduces and oxidizes the substance from which it accepts electron. So, it acts as oxidising agent. 

Test: Classification of Elements & Periodicity in Properties - 1 - Question 20

The sequence of ionic mobility in aqueous solution is 

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 20

Smaller the size of cation, higher will be the hydration and its effective size will increase and hence mobility in aqueous solution will decrease. Larger size ions have more ionic mobility due to less hydration. Thus the degree of hydration of M+ ions decreases from Li+ to Cs+. Consequently the radii of the hydrated ion decreases from Li+ to Cs+. Hence the ionic conductance of these hydrated ions increases from Li+ to Cs+ 

Test: Classification of Elements & Periodicity in Properties - 1 - Question 21

In the periodic table Electronegativity generally

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 21

electronegativity increases on moving from left to right while decreases as we move down the group.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 22

Al(3+)<Mg(2+)<Na+<F−<O(2−)<N(3−) The above can be aptly described as

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 22

They are isoelectronic species with same number of electrons. Higher is the atomic number, higher will be the effective nuclear charge, lower will be the size.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 23

Generally, the first ionization energy increases along a period. But there are some exceptions. One which is NOT an exception is ______.

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 23

Sodium (Na) has a lower ionisation energy than Magnesium (Mg). The reasons for this are as follows:

  • The removal of an electron from Na requires less energy.
  • This is due to Na's lower effective nuclear charge.
  • Generally, ionisation energy increases across a period.

Therefore, this trend indicates that Na is not an exception to the rule.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 24

Which of the following will have the most negative electron gain enthalpy and which one the least negative? P, S, Cl, F.

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 24

Cl just need one more electrons to get fully filled electronic configuration so it has most negative electron gain enthalpy while P has d3 configuration which is already stable so it cannot accept one electron extra easily so have large positive electron gain enthalpy.

Test: Classification of Elements & Periodicity in Properties - 1 - Question 25

For Atomic numbers greater than 100 IUPAC has made recommendation that until a new element ‘s discovery is proved, and its name is officially recognized, a systematic nomenclature be derived directly from the atomic number of the element using the numerical roots for 0 and numbers 1-9. The name 'sept' in this scheme corresponds to the digit

Detailed Solution for Test: Classification of Elements & Periodicity in Properties - 1 - Question 25

sept is for 7.

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