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Classification of Elements & Periodicity in Properties Test 15

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Classification of Elements & Periodicity in Properties Test 15
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Why do noble gases have positive electron gain enthalpy?

    Solution

    Due to stable configuration they do not accept an electron and hence they have positive electron gain enthalpy.

  • Question 2
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    As we move from left to right, the electronegativity increases. An atom which is highly electronegative has

    Solution

    It is difficult to remove an electron from a highly electronegative element.

  • Question 3
    1 / -0

    Study the given diagram and fill up the blanks with appropriate choice.
    925938

    Solution

    Ionisation enthalpy, electron gain enthalpy and electronegativity increases in a period from left to right while atomic radius decreases.

  • Question 4
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    Fill in the blanks with the appropriate option. 
    The ability of an atom to attract shared electrons to itself is called (i). It is generally measured on the (ii) scale. An arbitrary value of (iii) is assigned to fluorine (have greatest ability to attract electrons). It generally (iv) across a period and (v) down a group.

    Solution

    Electronegativity is measured on Pauling scale. Fluorine, the most electronegative element is given the value 4.0. Electronegativity increases from left to right across a period while decreases down a group.

  • Question 5
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    In which of the following, the order is not in accordance with the property mentioned.

    Solution

    Electronegativity decreases down the group.

  • Question 6
    1 / -0

    Choose the incorrect statement.

    Solution

    Electronegativity increases across a period.

  • Question 7
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    In the periodic table, the maximum chemical reactivity is at the extreme left (alkali metals) and extreme right (halogens). Which properties of these two groups are responsible for this?

    Solution

    Elements on the left side have lowest ionisation enthalpy due to which they can very easily lose electrons while the elements onthe right can accept electrons easily as they show highest negative electron gain enthalpy.

  • Question 8
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    An element X has atomic number 19. What will be the formula of its oxide?

    Solution

    Element X has atomic number 19. Its valency will be one. Hence, the formula ofits oxide will be X2O.
    Z = 19; 1s22s22p63s23p64s1

  • Question 9
    1 / -0

    An element 'P' has atomic number 56. What will be the formula of its halide?

    Solution

    Z = 56; [Xe] 6s2
    Formula of halide is PX2.

  • Question 10
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    Predict the formula of stable compound formed by an element with atomic number 114 and fluorine.

    Solution

    Atomic number 114 will fall into carbon family, hence it will be tetravalent. So, formula with F will be AF4.

  • Question 11
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    Predict the formula of a compound formed by aluminium and sulphur.

    Solution

    Al - 3e- → Al3+, S + 2e- → S2-
    The formula for its sulphide is AI2S3.

  • Question 12
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    There are many elements in the periodic table which exhibit variable valency. This is a particular characteristic of

    Solution

    Transition elements show variable valency due to the presence of vacant d-orbitals.

  • Question 13
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    Predict the formulae of the binary compounds formed by combination of the following pairs of elements:
    (i) Magnesium and nitrogen
    (ii) Silicon and oxygen

  • Question 14
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    The first element of the groups 1 and 2 are different from other members of the respective groups. Their behaviour is more similar to the second element of the following groups. What is this relationship known as?

    Solution

    The relation between 1st element of a group and 2nd element of the next group is called diagonal relationship.

  • Question 15
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    The main reason for showing anomalous properties of the first member of a group in s or p-block is

    Solution

    Due to small size, large charge/radius ratio and high electronegativity, the first elementshows anomalous behaviour.

  • Question 16
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    When we go from left to right in a period,

    Solution

    The basic nature of the oxides decreases from left to right in a period, e.g., in third period.

  • Question 17
    1 / -0

    Which of the following oxides is neutral in nature?

    Solution

    CO isa neutral oxide. SrO isa basic oxide. AI2O3 is an amphoteric oxide while CO2 is an acidic oxide.

  • Question 18
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    What is the common property of the oxides CO, NO and N2O?

    Solution

    CO, NO and N2O are neutral oxides.

  • Question 19
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    The correct order of acidic character of oxides in third period of periodic table is

    Solution

    Acidic character of oxides increases as we move from left to right in a period.

  • Question 20
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    Which of the following sets of oxides is amphoteric in nature?

    Solution

    AI2O3, As2O3 and ZnO are amphoteric in nature.

  • Question 21
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    (A), (B) and (C) are elements in the third short period. Oxide of (A) is ionic, that of (B). is amphoteric and of (C) a giant molecule. (A), (B) and (C) have atomic number in the order ;

    Solution

    A, B and C are magnesium, aluminium and sillicon. Magnesium forms ionic oxide, (MgO) aluminium forms amphoteric oxide, (AI2O3) and silicon forms a giant molecule (SiO2).

  • Question 22
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    What is common between given cations and anions, O2-, F-, Na+, Mg2+, Al3+?

  • Question 23
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    Which of the following is arranged in order of increasing metallic character?

    Solution

    Metallic character increases down a group and decreases along a period as we move from leftto right. Hence, the order of increasing metallic character is :
    P < Si < Be < Mg < Na

  • Question 24
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    Which is the most electropositive element?

    Solution

    The element that belongs to group 1 is the most electropositive. Electropositive character increases down the group.

  • Question 25
    1 / -0

    Among the elements with atomic numbers 9, 12, 16 and 36 which is highly electropositive?

    Solution

    Z=9; 1s22s22p5
    Z= 12; 1s22s22p63s2
    Z =16; 1s2 2s2 2p6 3s2 3p4
    Z = 36: 1s2s2 2p6 3s2 3p63d10 4s24p6
    Electropositivity decreases along the period while it increases going down the group.

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