शास्त्रीय कारणे लिहा (कोणतीही तीन) :
एका गणसमूहातील मूलद्रव्यांची संयुजता समान असते.
शास्त्रीय कारणे लिहा (कोणतीही तीन) :
एका गणसमूहातील मूलद्रव्यांची संयुजता समान असते.
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Solution and Explanation
Step 1: संकल्पना समजून घेणे.
आवर्त सारणीमध्ये एकाच गणसमूहात (group) येणाऱ्या मूलद्रव्यांची बाह्य इलेक्ट्रॉन व्यवस्था समान असते. बाह्य कक्षेतील इलेक्ट्रॉन्सच्या संख्येमुळे संयुजता ठरते.
Step 2: उदाहरण.
उदा. पहिल्या गणसमूहातील मूलद्रव्ये (Li, Na, K, Rb) यांमध्ये प्रत्येकाच्या बाह्य कक्षेत एकच इलेक्ट्रॉन असतो, त्यामुळे सर्वांची संयुजता 1 असते.
Step 3: निष्कर्ष.
एका गणसमूहातील सर्व मूलद्रव्यांची बाह्य इलेक्ट्रॉन रचना समान असल्याने त्यांची संयुजता देखील समान असते.
Top Questions on Periodic Classification of Elements
- Choose the correct alternative:
Alkaline earth metals have valency 2. This means that their position in the modern periodic table is in .................... .- Maharashtra Class X Board - 2025
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Questions Asked in Maharashtra Class X Board exam
- Appreciation of the poem:
Read the following poem and write an appreciation of it with the help of the points given below:
Stopping by Woods on a Snowy Evening
Whose woods these are I think I know.
His house is in the village, though;
He will not see me stopping here
To watch his woods fill up with snow
My little horse must think it queer
To stop without a farmhouse near
Between the woods and frozen lake
The darkest evening of the year.
He gives his harness bells a shake
To ask if there is some mistake.
The only other sound’s the sweep
Of easy wind and downy flake.
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
Robert Frost
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- Reading Comprehension
- Pick out the infinitive from the following sentence :
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- Out of the following which is a Pythagorean triplet ?
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\).
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- Properties of Triangles
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.