Self Studies
Self Studies
Self Studies
Self Studies

D and F - Block Elements Test - 9

Result Self Studies

D and F - Block Elements Test - 9
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    4 / -1

     

    Which one of the following is an electronic configuration of thorium?

    Solution

     

     

    Thorium (Z = 90) have [Rn] 5f06d27s2 configuration.

     

     

  • Question 2
    4 / -1

    Which of the following is general electronic configuration of actinides?

    Solution

    General EC = [Rn] 5f1-146d0-17s2.

  • Question 3
    4 / -1

    The actinoids Exhibit more member of oxidation states in general than the lanthanoids. This is because

    Solution

    The wide range of oxidation state of actinoids is due to the fact that the 5f, 6d and 7s energy levels are of comparable energies and all these three subshells can participate in comparison to lanthanoid. The 5f orbitals extend farther from the nucleus than the 4f orbitals.

  • Question 4
    4 / -1

    Identify the incorrect statement among the following.

    Solution

    4f electrons have greater shielding effect but 5f electrons have poor shielding effect.

  • Question 5
    4 / -1

    In which 5f subshell is half-filled?

    Solution

    Americium (2 = 95) and curium (Cm, Z = 96) have [Rn] 5f76d07s2 and [Rn] 5f76d17s2 configuration, respectively.

  • Question 6
    4 / -1

    Actinides

    Solution

    Actinides are highly electropositive and reactive metals.

  • Question 7
    4 / -1

    Which of the following is not an actinide?

    Solution

    Actinoids are from thorium (Z = 90)to Lawrencium (Z = 103) in the Periodic table So, rutherfordium (Z = 104) is not an actinoid.

  • Question 8
    4 / -1

    The compound used in enrichment of uranium for nuclear power plant is

    Solution

    UF6 is used in enrichment of uranium for nuclear power plant.

  • Question 9
    4 / -1

    Which one of the following shows oxidation state upto + 7?

    Solution

    Plutonium (Z = 94) have [Rn] 5f66d07s2 configuration and show oxidation state upto +7.

  • Question 10
    4 / -1

    The general electronic configuration of actinides is (n - 2) f1-14 (n - 1) d0-1 ns2.

    Which one of the following actinides has one electron in 6d-orbital?

    Solution

    Cm(Z = 96)have [Rn] 5f76d17s2,
    Lr (Z = 103) have [Rn] 5f146d17s2
    Np (Z = 93) have [Rn] 5f46d17s2

  • Question 11
    4 / -1

     

    Which one of the following is a diamagnetic ion?

    Solution

    Co+2 = [Ar] 3d7
    Cu2+ = [Ar] 3d9
    Mn+2 = [Ar] 3d5
    Sc+3 = [Ar]
    We can see that only Sc+3  has no unpaired electron, so it is a diamagnetic ion.

     

  • Question 12
    4 / -1

     

    The elements which lie between s and p block elements in the long form periodic table are called as:

     

  • Question 13
    4 / -1

     

    Which form of silver is colourless?

     

    Solution

     

    Silver in the form of Ag+ is colourless. For transition metal ions to exhibit color, their metal ions must have incompletely filled (n-1)d orbitals.

    Ag+ =4d10,5s0

    Ag+ has completely filled d orbitals hence is colourless.

  • Question 14
    4 / -1

     

    Melting point of d block elements across a period:

    Solution

     

    Till d5 unpaired electron are increasing so maximum bonding as unpaired electron take part in bonding. So melting point increases till d5 then start decreasing due to pairing which decreases metallic bond strength and thus melting point.

  • Question 15
    4 / -1

     

    The first ionization energy of the d-block elements are?

    Solution

     

    The first ionization energy of the d-block elements are between s and p-block elements. Thus they are more electropositive than p-block elements and less electropositive than s-block elements.

  • Question 16
    4 / -1

    Comprehension Type

    Direction (Q. Nos. 16 and 17) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

    Passage

    Rutherford and Soddy found that every uranium ore that they examined contained minute quantities of radium and an appreciable amount of non-radioactive lead. From this, they concluded that uranium is the starting substance in a series of radioactive transformations which pass through radium and ultimately ends with stable lead. These transformations are accompanied by spontaneous emissions of alpha and beta particles resulting in the formation of new atoms. All the nuclei from the initial element to the final stable element constitute a series called disintegration series. Mass number changes by 4 units per each alpha particle emission.
    There are four disintegration series. They are

    Q. 

    The number of alpha and beta particles emitted in the nuclear reaction  are

    Solution

    Number of alpha particles emitted

    Number of beta particles liberated
    = Double the alpha particles - change in atomic number
    = 8 - 7 = 1

  • Question 17
    4 / -1

    Rutherford and Soddy found that every uranium ore that they examined contained minute quantities of radium and an appreciable amount of non-radioactive lead. From this, they concluded that uranium is the starting substance in a series of radioactive transformations which pass through radium and ultimately ends with stable lead. These transformations are accompanied by spontaneous emissions of alpha and beta particles resulting in the formation of new atoms. All the nuclei from the initial element to the final stable element constitute a series called disintegration series. Mass number changes by 4 units per each alpha particle emission.
    There are four disintegration series. They are

    Q. 

    Lead isotope is the end product in all disintegration series except 

    Solution

    The end product of neptunium series is bismuth-209.

  • Question 18
    4 / -1

    Matching List Type

    Direction (Q. No. 18) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

    Solution

    (i)→ (q), (ii) → (s), (iii) → (r) , (iv) → (p)

  • Question 19
    4 / -1

     

    Direction (Q. Nos. 19-23) This section contains 5 questions. When worked out wiil result in an integer from 0 to 9 (both inclusive).

    Q. The number of electrons present in 5f subshell of berkelium are

    Solution

     

    Berkelium (Z = 97) : [Rn] 5f96d07s2

     

  • Question 20
    4 / -1

     

    The number of electrons present in 5f subshell of Th4+ ion are

     

    Solution

     

    Thorium +4 configuration (Z = 90) = [Rn] 5f06d07s0

     

  • Question 21
    4 / -1

     

    Number of elements that show + 6 oxidation state among Th, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm are

     

    Solution

     

    Only U, Np, Pu and Am show +6 oxidation state.

     

  • Question 22
    4 / -1

     

    The number of unpaired electrons in the element with atomic number 100 are

     

    Solution

     

    The element with atomic number 100 is fermium. Its configuration is [Rn] 5f126d07s2.

     

  • Question 23
    4 / -1

     

    Pb, in this transformation, number of alpha particles (α) liberated

     

    Solution

     

    Number of alpha particles (α) liberated = change in mass number/4 = 238 - 206/4 = 8

  • Question 24
    4 / -1

     

    Statement Type

    Direction (Q. Nos. 24 and 25) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

    Q. 

    Statement I : Actinides show a large number of oxidation state whereas lanthanides show a limited number of oxidation state.

    Statement Il : Energy gap between 4f, 5d and 6s subshells is small whereas that between 5f, 6d and 7s subshell is large.

     

    Solution

     

    statement ll is wrong

    The energy gap among 4f, 5d and 6s is large where as the energy gap among 5f, 6d and 7s is small.

  • Question 25
    4 / -1

    Statement I : Actinides form relatively less stable complexes as compared to lanthanides.

    Statement II : Actinides can utilise their 5f -orbitals along with 6d-orbitals in bonding but lanthanides do not use their 4f-orbitals for bonding.

    Solution

    Acitinides form relatively more stable complexes as compared to lanthanides.

Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now