The pair that has similar atomic radii is :
- Sc and Ni
- Ti and HF
- Mo and W
- Mn and Re
The Correct Option is C
Solution and Explanation
Top Questions on The Lanthanoids
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Questions Asked in JEE Main exam
- Let \( M \) denote the set of all real matrices of order 3 x 3 and let \( S = \{-3, -2, -1, 1, 2\} \). Let
\( S_1 = \{A = [a_{ij}] \in M : A = A^T \text{ and } a_{ij} \in S, \forall i, j\} \),
\( S_2 = \{A = [a_{ij}] \in M : A = -A^T \text{ and } a_{ij} \in S, \forall i, j\} \),
\( S_3 = \{A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j\} \).
If \(n(S_1 \cup S_2 \cup S_3) = 125\), then \( \alpha \) equals:- JEE Main - 2025
- Linear Algebra
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
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JEE Main Notification
GMC Azamgarh Admission 2025: Dates, Courses, Fees, Eligibility, SelectJuly 22, 2025Concepts Used:
Lanthanoids
Lanthanoids are at the top of these two-row, while actinoids are at the bottom row.
Properties of Lanthanoids
Lanthanoids are inclusive of 14 elements, with atomic numbers 58-71:
- Cerium - Xe 4f1 5d1 6s2
- Praseodymium - Xe 4f3 6s2
- Neodymium - Xe 4f4 6s2
- Promethium - Xe 4f5 6s2
- Samarium - Xe 4f6 6s2
- Europium - Xe 4f7 6s2
- Gadolinium - Xe 4f7 5d1 6s2
- Terbium - Xe 4f9 6s2
- Dysprosium - Xe 4f10 6s2
- Holmium - Xe 4f11 6s2
- Erbium - Xe 4f12 6s2
- Thulium - Xe 4f13 6s2
- Ytterbium - Xe 4f14 6s2
- Lutetium - Xe 4f14 5d1 6s2
These elements are also called rare earth elements. They are found naturally on the earth, and they're all radioactively stable except promethium, which is radioactive. A trend is one of the interesting properties of the lanthanoid elements, called lanthanide contraction.