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Dobereiner's triads - class-XI

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According to the law of triads :

  1. the properties of the middle element were in between those of the other two members.

  2. three elements arranged according to increasing weights have similar properties.

  3. the elements can be grouped in the groups of six elements.

  4. every third element resembles the first element in periodic table.


Correct Option: A
Explanation:

According to Dobereiner's law of triads, the properties of the middle element were in between those of the other two members. The elements were arranged in a group of three (called triads). The atomic weight of a middle element was almost the average of the other two. Three elements of the triad had similar chemical properties.

The law of triad is applicable to a group of :

  1. $Cl, Br, I$

  2. $C, N, O$

  3. $Na, K, Rb$

  4. $H, O, N$


Correct Option: A,B
Explanation:

For Dobereiner's triads, the atomic weight of middle element is almost average of other two. Thus, the law of triad is applicable to (1) $Cl, Br, I$ and (2) $C, N, O$. However, it is not applicable to (3) $Na, K, Rb$ and (4) $H, O, N$. 


(1) For $Cl, Br$ and $I$, $\displaystyle \dfrac {35.5+127}{2} = 81.25 \simeq 80$ 


(2) For $C, N, O$ $\displaystyle \dfrac {12+16}{2}=14$ 

(3) For $Na, K, Rb$ $\displaystyle \dfrac {23+85.5}{2}=54.25 \neq 39$ 

(4) For $H,O,N$ $\displaystyle \dfrac {1+14}{2}=7.5 \neq 14$

In Doberenier triads, the atomic weights of the elements in a triad are in increasing order.

  1. True

  2. False


Correct Option: A
Explanation:

True, atomic weight of the element in a Doberenier traid is in increasing order. Example: LI, Na, and K with atomic weight 6, 23, and 39 respectively.

Example of Dobereiner's triad is _________.

  1. $Li , Al , Ca$

  2. $Li , Na , K$

  3. $Li , K , Na$

  4. $K , Al , Ca$


Correct Option: B
Explanation:

$Li,Na,K$ is the example of Dobereineer's traid. Here, average of atomic weights of  'Li' and 'K' equals atomic weight of 'Na'.


$\dfrac{Weight\ of\ Li+ weight\ of\ K}{2}$ =$\dfrac{7+39}{2}$ =$\dfrac{46}{2}$ =23


=weight of $Na$.

The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. Such an arrangement of elements called as :

  1. Dobereiner triad.

  2. Mendeleev triad.

  3. Modern triad.

  4. none of the above.


Correct Option: A
Explanation:

Dobereiner observed in few atoms that mass of one atom is equal to the average of two atoms. He defined almost all properties basing on atomic weight.Later on it was proved wrong any-ways. Hence, this is called Dobereiner triad. Lithium has 7, sodium has 23 where as potassium has 39 as atomic masses.
$ 23 = (7+39)/2 = 23$

Here, we could observe that sodium atomic mass is average of potassium and lithium masses. 

The law of triads was given by :

  1. Dobereiner

  2. Lewis

  3. Mendeleev

  4. Dalton


Correct Option: A
Explanation:

In the year 1829, Johann Wolfgang Dobereiner, a German scientist, was the first to classify elements into groups based on John Dalton's assertions. He grouped the elements with similar chemical properties into clusters of three called 'Triads'. The distinctive feature of a triad was the atomic mass of the middle element. When elements were arranged in order of their increasing atomic mass, the atomic mass of the middle element was approximately the arithmetic mean of the other two elements of the triad.

If the two members of a Dobereiner's triad are chlorine and iodine, then the third member of this triad is :

  1. fluorine

  2. bromine

  3. sodium

  4. calcium


Correct Option: B
Explanation:

According to Dobereiner, the atomic mass of the middle element is nearly equal to the arithmetic mean of the other two. Therefore, the atomic mass of the third element is $=$ $\dfrac{{35.5}+{127}}{2}$ $=$ 81.25, which is the atomic mass of bromine.

If the two members of a Dobereiner triad are phosphorus and antimony, the third member of this triad is :

  1. arsenic

  2. sulphur

  3. iodine

  4. calcium


Correct Option: A
Explanation:

According to Dobereiner, the atomic mass of the middle element was nearly equal to the arithmetic mean of the other two. Therefore, the atomic mass of the third element is $=$ $\frac{{31}+{121.75}}{2}$ $=$ 76.37 which is nearly equal to the atomic mass of arsenic.

Cl, Br, I  follows Dobereiner's triads, the atomic masses of Cl and I are 35.5 and 127 respectively. The atomic mass of Br is : 

  1. 162.5

  2. 91.5

  3. 81.25

  4. 45.625


Correct Option: C
Explanation:

According to Dobereneir's triads the atomic mass of Br will be average of the atomic masses of CI & I
 $=$ $\frac{{35.5}+{127}}{2}$ $=$ 81.25

Which of the following sets of atomic masses do not correspond to Dobereiner's triad?

  1. 7, 23, 39

  2. 20, 38, 56

  3. 47, 91, 178

  4. 54, 95, 183


Correct Option: C,D
Explanation:

The atomic mass of the middle element should be the average of the atomic masses of the first and third elements. In case of C and D options, the atomic masses do not correspond to that rule.

Which of the following triads does not follow Dobereiner's law of triads?

  1. Li, Na, K

  2. Ca, Sr, Ba

  3. Be, Mg, Ca

  4. Cu, Ag, Au


Correct Option: D
Explanation:

Since the atomic weight of Ag is not equal to the mean of the atomic weights of Cu and Au.


Thus, (Cu, Ag, Au) triad do not follow Dobereiner's law of triads.

Option D is correct.

Which of the following set of elements cannot be a triad?

  1. $Li, Na, K$

  2. $B, Al, Ga$

  3. $Be, Mg, Ca$

  4. $Cl, Br, I$


Correct Option: B
Explanation:

Triad can be only when the middle element is having mean atomic weight of other two elements $,$ which $B, Al, Ga$ does not have such triad$.$

Which of the following elements follow Dobereiner's law of Triads?

  1. Na = $23$, Li = $7$ and K = $39$

  2. Cl = $35$, Br = $80$ and F = $19$

  3. Ca = $40$, Mg = $24$ and Sr = $88$

  4. Na = $23$, K = $39$ and Rb = $85$


Correct Option: A
Explanation:

According to Dobereiner's, the atomic weight of the middle element is nearly the same as the average of the atomic weights of the other two elements.


$Li(7)+K(39)\longrightarrow \cfrac{7+39}{2}=Na(23)$

So, $Na=23, Li=7$ and $K=39$ 

The correct option is ( A ) 

Following triads have approximately equal size

  1. ${ Na }^{ + },\quad { Mg }^{ 2+ },\quad { Al }^{ 3+ }$

  2. ${ F }^{ - },\quad Ne,\quad { O }^{ 2- }$

  3. $Fe,\quad Co,\quad Ni$

  4. ${ Mn }^{ + },\quad { Fe }^{ 2+ },\quad Cr$


Correct Option: C

The law of  traids is not applicable on

  1. $Cl, Br,I$

  2. $Na, K, Rb$

  3. $S, Se, Te$

  4. $Ca, Sr, Ba$


Correct Option: B

The law of traids is applicable to :

  1. $C, N, O$

  2. $H, O, N$

  3. $Na, K, Rb$

  4. $Cl, Br, I$


Correct Option: D

In the Doberiener's triad, all three elements have similar:

  1. electronic configuration

  2. properties

  3. number of shells

  4. both A and B


Correct Option: B
Explanation:

In the Doberiener's triad, all three elements have similar properties and the atomic weight of the middle member of each triad is very close to the arithmetic mean (average) of the other two elements.


The Dobereiner's triads include (1) Li, Na, K (2) Ca, Sr, Ba and (3) Cl, Br, I.

Hence, the correct option is $\text{B}$

The law of triads is not applicable to :

  1. $Os, Ir, Pt$

  2. $Ca, Sr, Ba$

  3. $Fe, Co, Ni$

  4. $Ru, Rh, Pt$


Correct Option: C,D
Explanation:

The law of triads does not apply to $Ru, Rh, Pt$.


According to the law of triads, the atomic weight of the middle element is approximately equal to the arithmetic mean of the other two elements.

The atomic weights of $Ru, Rh$ and $Pt$ are $101, 102.9$ and $195$ respectively.

The arithmetic mean of the atomic weights of $Ru$ and $Pt$ is $\displaystyle \dfrac {101+195}{2} = 148$
Hence, the atomic weight of $Rh$ is not equal to the arithmetic mean of the atomic weights of $Ru$ and $Pt$.

$Fe = 56, Co = 59$ and $Ni = 59$

Mean of the weights of $Fe$ and $N$i is not equal to the weight of $Co$.


Hence, the correct options are $\text{C}$ and $\text{D}$

Atomic wt. of P is 31 and Sb is 120. What will be the atomic wt. of As, as per Dobernier triad rule?

  1. $151$

  2. $75.5$

  3. $89.5$

  4. Unpredictable


Correct Option: B
Explanation:

Atomic wt. of P is 31 and Sb is 120. The atomic wt. of As, as per Dobernier triad rule will be average of two. It will be $\displaystyle \frac {31+120}{2} = \frac {151}{2} =75.5 $

Atomic wt. of $Cl = 35.5$ and of $I = 127$. According to Doeberiner triad rule atomic wt. of $Br$ will be:

  1. $80.0$

  2. $162.5$

  3. $81.25$

  4. $91.5$


Correct Option: C
Explanation:

Atomic wt. of Cl = 35.5 and I = 127.


According to Doberiner triad rule:

Br =$\dfrac {35.5 +127}{2}=\dfrac {162.5}{2}=81.25$

Hence, the correct option is $\text{C}$

If three elements X, Y, and Z form a Dobereiner's triad and atomic weights of X and Z are 9 and 40 respectively, then the atomic weight of the element Y is approximately:

  1. 24.5

  2. 49

  3. 34.5

  4. 29


Correct Option: A
Explanation:

According to Dobereiner, the atomic weight of the central element of the triad is the arithmetic mean of the atomic weights of the other two members. 


Therefore, atomic weight of Y = $\cfrac{9 + 40}{2} = 24.5\ g$


Hence, option A is correct.

Law of Triad was proposed by:

  1. Newland

  2. Gay Lussac

  3. Mendeleev

  4. Dobereiner


Correct Option: D
Explanation:

Dobereiner, in 1817 suggested a relationship between the properties of elements and their atomic weights. According to Dobereiner, the atomic weight of the middle element is nearly the same as average of the atomic weights of other two elements.

The element in between lithium and potassium in Dobereiner's classification is:

  1. Mg

  2. Na

  3. Ca

  4. Rb


Correct Option: B
Explanation:

Li, Na, K formed a Dobereiner's triad. Thus the element between Lithium and potassium is Sodium (Na).

Chlorine, Y and Iodine form a Dobereiner's triad. Identify the atomic weight of Y.

  1. 162.5

  2. 81.25

  3. 121.5

  4. 90.5


Correct Option: B
Explanation:
Cl, Y and I for Dobereiner's triad. Thus weight of Y is arrange of weight of Cl and I
$\therefore$ Atomic weight of Y is $\cfrac { 35.5+127 }{ 2 } =81.25$

Three elements X, Y and Z form a Dobereiner triad. The ratio of the atomic weight of X to that of Z is 7: 25.


 If the sum of the atomic weights of X and Z is 160, find the atomic weights of X, Y and Z.

  1. X $\rightarrow$ 35, Y $\rightarrow$ 80, Z $\rightarrow$ 125

  2. X $\rightarrow$ 125, Y $\rightarrow$ 80, Z $\rightarrow$ 35

  3. X $\rightarrow$ 80, Y $\rightarrow$ 35, Z $\rightarrow$ 125

  4. X $\rightarrow$ 80, Y $\rightarrow$ 125, Z $\rightarrow$ 35


Correct Option: A
Explanation:
Let weight of X be $7a$  $\therefore$ Atomic weight of $Z=25a$

$\therefore7a+25a=160$

$\Rightarrow a=5$

Weight of X$=35$ and of Z is $125$

Atomic weight of Y is arranged of X and $2=\cfrac { 35+125 }{ 2 } =80$

Hence, the correct option is $\text{A}$

When a Dobereiner triad is considered, the sum of atomic weights of extreme elements X and Z is 177.6 and difference of Z and Y is five times the number of protons present in a neon atom. Identify X, Y and Z.

  1. $X \rightarrow Ca, Y \rightarrow Sr, Z \rightarrow Ba$

  2. $X \rightarrow Li, Y \rightarrow Na, Z \rightarrow K$

  3. $X \rightarrow K, Y \rightarrow Rb, Z \rightarrow Cs$

  4. $X \rightarrow Sr, Y \rightarrow Ba, Z \rightarrow Ra$


Correct Option: A
Explanation:
The atomic weight of Y is avergae of X and Z
$\therefore { M } _{ Y }=\cfrac { 177.6 }{ 2 } =88.8$
$88.8$ is weight of Sr. So Y is Sr. According to this only option A is correct.

Which of the following sets of atomic masses do not correspond to Dobernier traid?

  1. $7, 23, 39$

  2. $20, 38, 56$

  3. $47, 91, 178$

  4. $54, 95, 183$


Correct Option: C,D
Explanation:

The following sets of atomic masses do not correspond to Dobernier traid.
(C) $\displaystyle 47, 91, 178$
(D) $\displaystyle 54, 95, 183$
In these sets, the atomic mass of middle element is not equal to the mean of atomic masses of first and last element.
$\displaystyle 91 \neq \dfrac {47+178}{2}$
$\displaystyle 95 \neq \dfrac {54+183}{2} $

Dobereiner was the first person to illustrate the relationship between the ____________ of elements and their properties.

  1. atomic masses

  2. electron negativity

  3. molecular masses

  4. number of protons


Correct Option: A
Explanation:

Dobereiner grouped elements in set of triads. Thus each group had 3 elements. Dobereiner classified elements into groups based on their properties. He illustrate the relationship between the atomic masses of elements and their properties with his law of triads.

Which of the following statements is not true about Dobereiner's law of triads?

  1. The law was applicable to all known elements

  2. Nitrogen family does not obey the law of triads

  3. Dobereiner's classification was not very useful

  4. The atomic mass of the central element of each triad was merely the arithmetic mean of of the atomic masses of other two elements


Correct Option: A
Explanation:
Döbereiner’s triads could find only three triads; .i.e total of 9 elements only but there were many more elements known at that time.

Therefore Dobereiner’s could not classify most of the elements known at that time.

Hence, the correct option is $A$

In a triad of $A, B, C$ elements if the atomic masses of $A$ and $C$ respectively are $100$ and $200$, then the atomic mass of $B$ is:

  1. $300$

  2. $175$

  3. $125$

  4. $150$


Correct Option: D
Explanation:

According to Dobereiner's law of traid, the average of the masses of the first and the last element is equal to the atomic mass of the middle element. So, $\dfrac{100+200}{2}$ = 150. Hence, the atomic mass of B is 150.


Hence, the correct option is $\text{D}$

In Doberenier triads, the atomic weights of the elements in a triad are in increasing order.
State whether the given statement is true or false.

  1. True

  2. False


Correct Option: A
Explanation:

In 1829, Dobereiner proposed the Law of Triads: Middle element in the triad had the atomic weight that was the average of the other two members. Soon other scientists found chemical relationships extended beyond triads. Fluorine was added to Cl/Br/I group; sulfur, oxygen, selenium, and tellurium were grouped into a family; nitrogen, phosphorus, arsenic, antimony, and bismuth were classified as another group. In a group, those were in increasing order of atomic weight. Hence, the statement is true.

When a Dobereiner's triad is considered, the sum of atomic weights of the terminal elements $X$ and $Z$ is 162.5. The atomic weight of $Z$ is 50 units less than 5 times that of $X$. Identify the periods of $X, Y$ and $Z$ respectively in the present periodic table.

  1. $3,4,5$

  2. $4,5,6$

  3. $2,3,4$

  4. $3,5,4$


Correct Option: A
Explanation:

Let the atomic weights (in g/mol) of elements X, Y and Z be x, y and z respectively.
The atomic weight of Z is 50 units less than 5 times that of X.

$\displaystyle z = 5x-50$.....(1)

The sum of atomic weights of the terminal elements X and Z is 162.5
$\displaystyle x+z=162.5$
$\displaystyle z = 162.5-x$......(2)

Substitute (2) into (1)
$\displaystyle 162.5-x=5x-50$
$\displaystyle 162.5=6x-50$
$\displaystyle 6x=212.5$
$\displaystyle x=35.4$ g/mol

Substitute this in equation (2)
$\displaystyle z = 162.5-x = 162.5-35.4=127.1$ g/mol.

The atomic weight of Y is the arithmetic mean of the atomic weights of X and Z.
$\displaystyle y = \dfrac {x+z}{2}$
$\displaystyle y = \dfrac {35.4+127.1}{2}=81.3$ g/mol.

From the periodic table, we see that the elements with atomic weights 35.4, 81.3 and 127.1 belong to 3 (chlorine), 4 (bromine) and 5 (iodine).

Hence, the option (A) is the correct answer.

The early attempts to classify elements were based on:

  1. Atomic number

  2. Atomic mass

  3. Electronic configuration

  4. None,of these


Correct Option: B
Explanation:

In the periodic table of 1860 elements were arranged on the basis of their atomic weight properties. Elements with same chemical properties have nearly same atomic weight values.

Which of the following statements are true?

  1. All the trans-uranic elements are synthetic elements

  2. Elements of third group are called bridge elements

  3. Element of $1 s^2$ configuration is placed in IIA group

  4. Electronic configuration of elements of a group is same


Correct Option: A,B

Which is mismatched regarding the position of the elements as given below?

  1. X(Y=89), f - block, 6th period

  2. Y(Z=100), f - block, 7th period

  3. Z(Z=115), d - block, 7th period

  4. Both (1) & (3)


Correct Option: D

Electronic configuration of silver atom in ground state.

  1. $\left[ Kr \right] { 3d }^{ 10 }{ 4s }^{ 1 }$

  2. $\left[ Xe \right] { 4f }^{ 14 }{ 5d }^{ 10 }{ 6s }^{ 1 }$

  3. $\left[ Kr \right] { 4d }^{ 10 }{ 5s }^{ 1 }$

  4. $\left[ Kr \right] { 4d }^{ 9 }{ 5s }^{ 2 }$


Correct Option: C
Explanation:

The electronic configuration of $Ag$ in ground state is:$[Kr]4d^{10}5s^1$

Write True or false:

The symbol of gold is Ag.

  1. True

  2. False


Correct Option: B
Explanation:

Symbol for gold is Au.

So, option-B is correct.

Neptunium is named after the: 

  1. Planet Neptune

  2. Scientst Naptuna

  3. Place Naptunnom

  4. None of these


Correct Option: A
Explanation:

Neptunium is named after the planet Neptune.

Tungsten is obtained from the latin name:

  1. Wolfram

  2. Wolfire

  3. Wolfoam

  4. Wolfox


Correct Option: A
Explanation:

The symbol for tungsten, W, comes from tungsten's Latin name Wolfram. The Latin name is derived from the tungsten mineral wolframite.


Hence, the correct option is $\text{A}$

Fill in the blanks 
Zero group elements are known as _________.

  1. chalcogen

  2. inert gases

  3. halogens

  4. alkali metals


Correct Option: B
Explanation:

The $'0'$ group elements have complete octate in outermost shell. As a result they have stable configuration due to which, they do not react with other elements. Hence they are known as inert gases.

An element $X$ belongs to fourth period and fifteenth group of the periodic table. Which of the following statements is true?

  1. It has a completely filled s-orbital and a partially filled d-orbital.

  2. It has completely filled s-and p-orbitals and a partially filled d-orbital.

  3. It has completely filled s-and p-orbitals and a half-filled d-orbital.

  4. It has a half-filled p-orbital and completely filled s-and d-orbitals.


Correct Option: D
Explanation:
Group 15 has a valence shell electronic configuration of $ns^2np^5$ and period $=4$$ means its $n=4$

thus electronic configuration of element $X$  is $4s^24p^3$

Elements of Group 15 and period 2 onwards are: $N, P, As, Sb, Bi$

Thus $X$ is Arsenic ($As$) and has electronic configuration of: $1s^22s^22p^63s^23p^63d^{10}4s^24p^3$. 

Thus, $s$ and $d-$orbitals are fully-filled and $p-$orbital is half-filled.

Hence, the correct option is $\text{D}$

Which of the following is an artificial element made by man?

  1. Gallium

  2. Iron

  3. Helium

  4. Americium


Correct Option: D
Explanation:
In chemistry, a synthetic element is a chemical element that does not occur naturally on Earth, and can only be created artificially. So far, 24 synthetic elements have been created (those with atomic numbers 95–118). Americium is one of them having atomic no. 95.

Hence, the answer is option $D$.
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