Tag: periodic table

Questions Related to periodic table

The element with the highest first ionisation potential is:

  1. Boron

  2. Carbon

  3. Nitrogen

  4. Oxygen


Correct Option: C

The ionisation potential of hydrogen atom is $13.6\ eV$. The energy of required to remove an electrons in the $n=2$ state of hydrogen atom is:

  1. $27.2\ eV$

  2. $13.6\ eV$

  3. $6.8\ eV$

  4. $3.4\ eV$


Correct Option: D
Explanation:
$E _n=-13.6\ ev\left (\dfrac {Z^2}{n^2}\right)=\dfrac {-E _1}{n^2}[I.E _1 =-E _1 =13.6\ eV \Rightarrow E _1=-13.6\ eV]$

$\therefore E _2 =\dfrac {-13.6\ ev}{4}=-3.4ev$

$\therefore I.E _2=-E _2 =3.4\ ev$.

Option D is correct.

In Mendeleev's periodic table few elements that are chemically similar are placed in the same groups.

  1. True

  2. False


Correct Option: A
Explanation:

Mendeleev arranged the elements having similar configuration placed in the vertical columns as they will have the same unpaired electrons hence reflecting the similar properties. In Mendeleev's periodic table few elements that are chemically similar are placed in separate groups.

The horizontal rows in Mendleev's table were called :

  1. groups

  2. rows

  3. periods

  4. none of the above


Correct Option: C
Explanation:

The horizontal rows in Mendleev's table were called periods. There are seven periods in Mendeleev's periodic table.

The vertical columns in Mendeleev's table were called :

  1. verticals

  2. periods

  3. columns

  4. groups


Correct Option: D
Explanation:

The vertical columns in Mendeleev's table were called groups. There are $8$ groups in Mendeleev's periodic table.

In Newland's classification, elements are arranged in an increasing order of their:

  1. mass number

  2. atomic number

  3. atomic weight

  4. all of these


Correct Option: C
Explanation:

Newlands arranged elements in increasing order of their atomic weight and stated that every eighth element resembled properties of first element. This is Newland's law of Octaves. Hence correct answer for asked question is option C-atomic weight.

Which of the following is the correct set of elements to Dobereiner's triads ?

  1. Li - 7, Na - 23, K -39

  2. Br - 80, Cl - 35.5, I - 127

  3. Fe - 55.85, Ni - 58.71, Co - 58.93

  4. All of these


Correct Option: A
Explanation:

Dobereiner's triads are a group of three elements in which the atomic mass of the middle element is equal to the average of atomic masses of the first and the third element.


For option A,
Average of atomic masses of first and last elements = $7+ \dfrac{39}{2} = 23$
This is the atomic mass of second element.

For option B.
Average of atomic masses of first and last elements = $80+ \dfrac{127}{2} = 103.5$
This is not the atomic mass of the second element.

For option C,
Average of atomic masses of first and last elements = $55.85+ \dfrac{58.93}{ 22} = 55.89$
This is not the atomic mass of the middle element.

Hence, the correct answer is option $A$.

Statement: In Dobernier classification of elements, the atomic weight of the middle element is equal to the average of the extreme elements of a triad.


State whether the given statement is true or false.

  1. True

  2. False


Correct Option: A
Explanation:

Dobereiner classified the elements into triads (group of three). He made the groups such that, the mass of middle element is the average of the mass of other $2$ elements. This is Dobernier's law of triads. 


Thus given statement is $\text{true}$.

The attempt for classifying elements by plotting the atomic masses of elements against the atomic volumes was made by:

  1. Dobereiner

  2. Newlands

  3. Lother Meyer

  4. Mendeleev


Correct Option: C
Explanation:

According to this if the atomic weights were plotted as ordinates and the atomic volumes as abscissae-the curve obtained a series of maxima and minima-the most electropositive elements appearing at the peaks of the curve in order of their atomic weights and electronegative element at the ascending positions.

The attempt for classifying elements by plotting the atomic masses of elements against the volumes was made by:

  1. Dobereiner

  2. Newland

  3. Lother Meyer

  4. Mendeleev


Correct Option: C
Explanation:

Lothar Meyer (1869) tried to classify by plotting a graph of atomic volume versus atomic masses of different elements. 


According to this if the atomic weights were plotted as ordinates and the atomic volumes as abscissae-the curve obtained a series of maxima and minima-the most electropositive elements appearing at the peaks of the curve in order of their atomic weights and electronegative element at the ascending positions.

Option C is correct.