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Classification of Elements and Periodicity in Properties Test - 9

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Classification of Elements and Periodicity in Properties Test - 9
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  • Question 1
    1 / -0

    Consider the isoelectronic species, \(Na ^+\) , \(Mg ^{2+}\) , \(F ^-\) and \(O ^{2-}\) . The correct order of increasing length of their radii is........

    Solution

    All of them are isoelectronic species. They have same number of electrons (10). Their radii would be different because of their different nuclear charges. The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size.

  • Question 2
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    Which of the following is not an actinoid?

    Solution

    Actinoids are elements with Z = 90-103. They are characterised by the outer electronic configuration (n-2) \(f ^{-14}\) (n-1) \(dns ^2\). The last electron added to each element is filled in f-orbital. Hence Terbium (Z = 65) is not an actinoid. It is a lanthanoid.

  • Question 3
    1 / -0

    Which of the following elements can show covalency greater than 4?

    (i) Be

    (ii) P

    (iii) S

    (iv) B

    Solution

    Both the above elements contain vacant d-orbitals that is why they can extend their covalency.

  • Question 4
    1 / -0

    Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionisation (i.e., absorb energy in the visible region of spectrum). The elements of which of the following groups will impart colour to the flame?

    (i) 2

    (ii) 13

    (iii) 1

    (iv) 17

    Solution

    Group I (alkali metals) and Group II elements (alkaline earth metals) are reactive metals with low ionization enthalpies.

  • Question 5
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    The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:

    Solution

    The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of ‘‘shielding” or “screening” of the valence electron from the nucleus by the intervening core electrons. For example, the 2s electron in lithium is shielded from the nucleus by the inner core of 1s electrons.

  • Question 6
    1 / -0

    In the following questions a statement of Assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.

    Assertion (A): Generally, ionisation enthalpy increases from left to right in a period.

    Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.

    Solution

    Ionization enthalpy depends upon two factors:

    (i) the attraction of electrons towards the nucleus, and

    (ii) the repulsion of electrons from each other.

  • Question 7
    1 / -0

    Match the correct atomic radius with the element.

    Element Atomic radius (pm)
    (i) Be (a) 74
    (ii) C (b) 88
    (iii) O (c) 111
    (iv) B (d) 77
    (v) N (e) 66
  • Question 8
    1 / -0

    The first ionisation enthalpies of Na, Mg, Al and Si are in the order:

    Solution

    Across a period, increasing nuclear charge outweighs the shielding. Consequently, the outermost electrons are held more and more tightly and the ionization enthalpy increases across a period. In case of Mg the outermost electron is in s-orbital that is a stable gas configuration and penetration effect of s-subshell is more as compared to Al in which outermost electron is in p-subshell.

  • Question 9
    1 / -0

    The electronic configuration of gadolinium (Atomic number 64) is

    Solution

    Gadolinium is a lanthanoid, it is a f-block element and the outer electronic configuration is (n - 2)\(f ^{-14}\) (n-1)\(d ^{0-1}\)\(ns ^2\).

  • Question 10
    1 / -0

    Which of the following sequences contain atomic numbers of only representative elements?

    (i) 3, 33, 53, 87

    (ii) 2, 10, 22, 36

    (iii) 7, 17, 25, 37, 48

    (iv) 9, 35, 51, 88

    Solution

    The p-Block elements comprise those belonging to Group 13 to 18 and these together with the s-Block elements are called the Representative Elements or Main Group Elements.

  • Question 11
    1 / -0

    The statement that is not correct for periodic classification of elements is:

    Solution

    According to Aufbau principle, 3d-orbitals are filled after 4s and before 4porbitals, as the energy of 4s is lesser than 3d.

  • Question 12
    1 / -0

    Match the correct ionisation enthalpies and electron gain enthalpies of the following elements

    Elements \(\Delta H_1\,\Delta H_2\,\Delta_{eg} H\)
    (i) Most reactive nonmetal (a) 419 3051 -48
    (ii) Most reactive metal (b) 1681 3374 -328
    (iii) Least reactive element (c) 738 1451 -40
    (iv) Metal forming binary halide (d) 2372 5251 +48
  • Question 13
    1 / -0

    Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is:

    Solution

    Within a group, electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Hence, the element with most negative electron gain enthalpy is chlorine.

  • Question 14
    1 / -0

    Which of the following elements will gain one electron more readily in comparison to other elements of their group?

    (i) S (g)

    (ii) Na (g)

    (iii) O (g)

    (iv) Cl (g)

    Solution

    For many elements energy is released when an electron is added to the atom and the electron gain enthalpy is negative. For example, group 17 elements (the halogens) have very high negative electron gain enthalpies because they can attain stable noble gas electronic configurations by picking up an electron.

  • Question 15
    1 / -0

    The period number in the long form of the periodic table is equal to

    Solution

    The period indicates the value of n for the outermost or valence shell. Hence, Period Number = maximum n of any element, where n refers to principal quantum number.

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